The equivalent capacity between A and B in the given circuit is 

$\left({\mathrm{C}}_1=4\mathrm{\mu F},{\mathrm{C}}_2=12\mathrm{\mu F},{\mathrm{C}}_3=8\mathrm{\mu F},{\mathrm{C}}_4=4\mathrm{\mu F} {\mathrm{C}}_5=8\mathrm{\mu F}\right)$

It is balanced wheastone’s bridge,
$\Rightarrow {\mathrm{C}}_2 \mathrm{ineffective} \Rightarrow {\mathrm{C}}_2\Rightarrow 0$
4, 8 are in series $\Rightarrow \frac{4\times 8}{12}=\frac{8}{3}$
4, 8 are in series $\Rightarrow \frac{4\times 18}{12}=\frac{8}{3}$
$\frac{8}{3},\frac{8}{3}$ are in parallel $\Rightarrow {\mathrm{C}}_{\mathrm{P}}=\frac{16}{3}\mathrm{\mu F}$