If you also route 8 buses at once without the pan pot, you have another 15K/8 = 1875 ohms.

Regarding panorama potentiometers. What will be with pan pot? How this fact will change bus impedance?

1883 module use 10K lin double pot with 2K2 resistor in parallel.

If we use pan pot it will give 5K in middle position (signal in centre pan) for both buses (for example if it's just two)

So we will have 15K bus resistor plus 1527Ohm (5K+2K2 in parallel) = 16K527. Right?

I did it without the pan pot because I thought it was probably the worst case. With the pan pot in it gets harder to work out because it is not obvious what pan pot position to use for worst case. If it is fully left or right, then only half the bus resistors are connected So you have 4 x 15K, 2 x 10K and 1 x 2.2K in parallel, which is 3.75K // 5K // 2.2k which is 2.14k //2.2k = 1086 ohms. So it is actually worse than 8 bus resistors alone.

With pan in the middle the top half of each pan pot is 5k // 2.2k = 1527 ohms

The bottom half of the pan pot is 4x15K // 5k = 3.75k // 5k = 2142 ohms

So each pot in total is 1527 + 2142 = 3669 so the load is half this or 1834 ohms - surprisingly close to the no pan pot case.

Regarding value of pan pot. Neve modules use double 10K pots with 2K2. What purposes of 2K2?

In some schematics I saw 47K pot or just single pot. Which way and why better?)

Neve used linear pots for pan pots because the two tracks are better matched. However, a linear pot would give a 6dB dip in the centre position (half way) when something nearer to about 3dB is required. At the centre the top half of the pot is 2.2k // 5k = 1527 ohms. With four buses routed the bottom half is 2142 so the attenuation is 2142/ (2142 +1527) = 0.5838 = 4.6dB. With one bus routed it would be 3750/(3750 + 1527) = 0.710 = 2.96 dB

Cheers

Ian