Calculate capacitance value and voltage rating in the event of ESD pulses
4 kV pulse from 150 pF in series with 330 ohms

The charge is 0.6 uC (as you said) and, ignoring the 330 ohm resistor (which limits the peak current), that charge becomes (after applying the pulse) distributed between the 150 pF and 2 series 47 nF capacitors. Given that the 150 pF is really small compared to 23.5 nF (the series capacitance of two 47 nF caps), you can assume (without much error) that the charge is all "adopted" by the 23.5 nF. 

This means that the voltage will be 25.53 volts. In other words twice the value you calculated because you need to regard the two 47 nF caps as being in series.

But, this doesn't accommodate the series diode and two more capacitors of 47 nF. In effect, the calculation above is valid for -4 kV. For +4 kV the capacitance is just 47 nF (ignoring the volt drop on the diode) because 2 series capacitors of 47 nF in parallel with an identical series connection of 2x 47 nF = 47 nF.

So, with a positive 4 kV pulse the voltage is 12.77 volts and, for -4 kV it is 25.53 volts.


  Is my approach correct?


Broadly speaking yes. And, more importantly, it is conservative in that we have assumed a worst case of the 330 ohm resistor being shorted and the diode (D0002) being shorted.

Rating the capacitors at 25 volts or above would be OK for this example.