Given, capacitance of each capacitor, $C=4 \mu \mathrm{F}$

The given circuit can be redrawn as shown in the figure:

Clearly, this is a balanced Wheatstone bridge circuit. So, we can remove the middle capacitor.

Now, the two $4 \mu \mathrm{F}$ capacitors in each of upper and lower branches are connected in series. So, equivalent capacitance in each branch is 

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{4}+\frac{1}{4}$

$\therefore C\text{'}=2 \mu \mathrm{F}$

Now as the capacitors are in parallel, so equivalent capacitance is

$C_{eq}=C_1+C_2=2+2$

$\Rightarrow C_{eq}=4 \mu \mathrm{F}$.