Rearranging the circuits, we get the following circuit
$\therefore $ equivalent capacitance between $A$ and $B$
$C_{A B}=\frac{4 \times 4}{4+4}=2 \,\mu F$
and equivalent capacitance between $C$ and $A$
$C_{C D}=\frac{8 \times 8}{8+8}=4 \, \mu F$
$\therefore C_{a b}=2 \, \mu F +4 \, \mu F =6 \, \mu F$