What is the purpose of this 5 V protection circuit?
Assuming that power is provided from the left (at VCC5V0) and taken from the right (at VCC5V0_ESD) I suppose the purpose of the circuit is to discharge C628 (and possibly more capacitors connected between VCC5V0_ESD and GND) and thus turn off the circuit to the right quickly as soon as power is removed.

I.e. Q8 will turn on only for a short moment after power is removed. 

Without transistors Q7A, Q7B, Q8 (e.g. just with diode D29 and capacitor C628) after removing power at VCC_5V0 voltage at VCC5V0_ESD might drop too slow leaving the circuit too long in an undefined/bad mode of operation.