The left and right sides of the P2 switches set the input and output common mode (relative to this block), respectively.  Let the voltage at the left branch be VCM1 and the other VCM2, then the charge equations come out to be the following.

(VOUTP+VOUTM)/2-VCM1=VCMFB-VCM2

There are times that the circuit as shown is useful if the desired voltage is VC, but as Tosei mentioned, the majority of times, VCM2 is a bias voltage for current mirrors.

As far as the resistors, I've never seen them used before, but I would have to agree with Tosei that they are for some extra pole.