i'm still trying to understand what you wrote. The new voltage readings seem wrong
OK, I'll walk through what I looked at.
Starting with the first transistor:
In the first schematic you posted, the emitter is at 6mV(?can't tell what character is in front of the 6). The base is at 2.9V. An operating transistor should have about 0.6V between base and emitter at low current, around 0.7V at higher current. Following the arrowhead on the schematic symbol, the base of the arrow should be more positive than the point of the arrow, so on a PNP transistor the emitter should be around 0.6V more positive than the base (or stated a different way, the base should be 0.6V more negative than the emitter).
In that first attempt the base was 2.3V more positive than the emitter, so obviously not biased into conduction.
In your operating example, the first transistor has an emitter voltage of -22.7mV, and a base voltage of -0.6V. The base is 0.58V more negative than the emitter, so it seems to be biased on at low quiescent current.
The current can be verified using Ohm's law, I=V/R. Since one end of the resistor connected to the emitter is at "ground" (0V reference node, I won't go into all the different meanings of "ground" here) the voltage across the resistor is the same as the emitter voltage, 22.7mV. 0.023/120=0.192mA.
Double checking if that is reasonable, you can see that the voltage across the collector resistor is -9V -(-6.96V) = -2.04V. The "-" isn't really important at this time, but for more complicated circuit analysis you want to keep track of the polarity of the voltage so that you can keep track of direction of current flow.
So 2.04V/10kOhms = 0.2mA.
There may be some slight inaccuracy there depending on whether you measured your power voltage and it really is 9.00V, or if you are using a battery and just wrote 9V there because it is around that (a brand new battery would probably be more like 9.3V or 9.5V, an old battery might be 8V).
But that is in the ball park considering that the collector current also has to supply base current to the transistor. A good rule of thumb if you don't know the transistor specifics is that the base current is probably around 1% of the emitter or collector current, so the base current is probably around 0.002mA.
That current goes through a 470k resistor from collector to base, which has -6.96V - (-0.6V) = -6.36V across it, so 6.36V/470kOhms=0.0013mA.
I will assume that you did not measure the exact battery voltage, so 0.0013mA is close enough to 0.0019mA or 0.002mA that I'll say everything matches up within the measurement accuracy and that the transistor is biased as expected.
For a circuit using NPN all the magnitudes would be similar, but the base would be more positive than the emitter, and the collector would be more positive than the base.
Just glancing at the second and third transistors they seem to be similar voltages, but higher voltage at the emitter indicating slightly higher bias current.
The last transistor has noticeably different voltages so I'll work through that and discuss why that might be the case:
Emitter is at -0.89V, base is at -1.4V, so 0.51V across the base-emitter junction. A little lower than the first transistor, so checking the emitter current to see if that makes sense: 0.89V/3300 Ohms = 0.27mA.
That is about the same as the others, so the transistor is probably operating normally. 0.51V is pretty low for an active base-emitter junction, possibly that transistor was noticeably hotter than the others when you measured, possibly that is just very low bias current for that model of transistor and so several mV variance would be expected. In any case it is biased about the same as the others, so makes sense.
The collector is at -4.9V, around 4.1V above the power supply, but with a 15k collector resistor instead of 10k. 4.1V/15k Ohms = 0.27mA, matches the emitter current, but in this case the base current comes directly from the power rail via a voltage divider, so the collector current will be almost the same as emitter current (emitter current will actually be a little over 1uA higher than collector current because it includes base current, but that is so small it is difficult to measure accurately).
A slight diversion to talk about that voltage divider providing the voltage to the base of the last transistor. That is a common biasing style, it is a little easier to analyze than having the base voltage derived from the collector voltage. With collector biasing you have to iterate through the voltages/currents to see where the final bias point stabilizes since the voltage at the collector depends on the current through the collector, which depends on the voltage at the base, which depends on the voltage at the collector. There is a circular dependency there that requires some iteration to figure out the correct resistor values.
With the voltage divider bias the voltage just depends on the power supply voltage and slightly on the base current. With a voltage divider you can calculate the voltage at the junction as power supply V * lower_resistor/(lower + upper resistors), so in this case -9V * 100K/(100K + 430K) = -1.698V.
There is an additional voltage drop across the top resistor because of base current. We know from the other transistors that 0.002mA is in the ballpark for base current, so 0.002m*430k=0.86V
That would be a base voltage of about -0.84V, and since the actual base voltage is -1.4V, in between the 1.70V calculated with no base current, and -0.84V with 2uA base current, that indicates that the actual base current is less than 2uA, which makes sense with a base-emitter voltage on the low side of normal.
If the base current is actually 1uA, the base voltage would be about -1.3V, so I will wave my hands a little bit and say that is around the right value.
I have not worked through the gains of all the stages to see what the expected amplitude will be at the output, but one thing to keep in mind is that the current delivered to the next device (probably a guitar amp?) has to come from Q1, so you could verify that the current determined by the output voltage swing and the input impedance of the next stage does not exceed around 0.1mA to 0.2mA, or the output transistor will go completely into cutoff.
Usually not a problem with guitar amps that usually have about 500k Ohm input impedance, just something to keep in mind.
