Effective capacitance between $A$ and $B$ in the figure shown is (all capacitances are in $\mu \mathrm{F}$ )

The points $C$ and $D$ will be at same potenitals since, $\frac{3}{6}=\frac{4}{8}$. Therefore, capacitance of $2 \mu \mathrm{~F}$ will be unaffected. So, the equivalent circuit can be shown as the effective capacitance in upper arm in series, is given by
$C_1=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \mu \mathrm{~F}$
The effective capacitance in lower arm in series, is given by $C_2=\frac{4 \times 8}{4+8}=\frac{32}{12}=\frac{8}{3} \mu \mathrm{~F}$
Hence, the resultant capacitance in parallel is given by $C=C_1+C_2=2+\frac{8}{3}=\frac{14}{3} \mu \mathrm{~F}$