d
The simplified circuit of the circuit given in question as follows:

The equivalent capacitance between \(C \& D\) capacitors of \(2\, \mu \mathrm{F}, 5\, \mu \mathrm{F}\) and \(5\, \mu \mathrm{F}\) are in parallel.

\(\therefore \mathrm{C}_{\mathrm{CD}}=2+5+5=12 \,\mu \mathrm{F}\)

(\(\because\) In parallel grouping \(C_{e q}=C_{1}+C_{2}+\ldots .+C_{n})\)

Similarly equivalent capacitance between \(\mathrm{E}\) \(\& \mathrm{BC}_{\mathrm{EB}}\)

\(=4+2=6\, \mu \mathrm{F}\)

Now equivalent capacitance between \(A \& B\)

\(\frac{1}{C_{e q}}=\frac{1}{6}+\frac{1}{12}+\frac{1}{6}=\frac{5}{12}\)

\(\Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{12}{5}=24\, \mu \mathrm{F}\)

(\(\because \) In series group-ing, \(\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} +  \ldots  \ldots  + \frac{1}{{{C_n}}}\)