In the circuit shown. The potential drop across each capacitor is (assuming the two diodes are ideal).
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Explanation

The diode `D_(1)` is reverse biased (open circuit), but the siode `D_(2)` is forward (short circuit).

`:.` The potential of the battery divides across the two capacitors in the inverse ratio of their capacities.
i.e, `(V_(1))/(V_(2)) = (C_(2))/(C_(1)) = (8)/(4) =(2)/(1)`
`V_(1) =(2)/(3) E = (2)/(3) xx 24 = 16 V`
`V_(2) = (1)/(3) E = (1)/(3) xx 24 = 8 V`.