PNP with zener diodes circuit
With 5V1 zeners, the base would be at 7V0 + the tiny voltage drop due to the base current. (Ib * 100)

Assuming a 0.7V Vbe drop the emitter will be at 7V7. This gives an emitter current of (12-7.7)/1000 = 4.3 mA

Taking beta = 100 the base current, Ib, will be 4.3/100 mA or 43uA

This takes the base voltage to 7V0 + 0.0043V which is approximately 7V0 (1dp).

Collector current will be Ie + Ib which is approx. Ie = 4.3mA, this gives a collector voltage of 4V3.

This is below the 5V1 of D2, so D2, R4 will have no effect.

If V1 increases over 7V the base voltage follows and the current through the transistor drops. The emitter voltage rises and the collector voltage falls.

If V1 decreases D1 clamps the base voltage at 6V9, the emitter voltage falls to 7V6 and the collector voltage rises to 4V4. 

This is still under the 5V1 of D2 so this part of the circuit still has no effect.

R4, D2 seem totally redundant.

What does the circuit actually do - no idea.