$C_{3}=\frac{2 \varepsilon_{0} k_{3} A}{3 d}, C_{4}=\frac{2 \varepsilon_{0} k_{4} A}{d}$

Given system of $C_{1}, C_{2}, C_{3}$ and $C_{4}$ can be simplified as $\therefore \frac{1}{C_{A B}}=\frac{1}{C_{1}+C_{2}+C_{3}}+\frac{1}{C_{4}}$

Suppose, $C_{A B}=\frac{k \varepsilon_{0} A}{d}$

$\quad \frac{1}{k\left(\frac{\varepsilon_{0} A}{d}\right)}=\frac{1}{\frac{2}{3} \frac{\varepsilon_{0} A}{d}\left(k_{1}+k_{2}+k_{3}\right)}+\frac{1}{\frac{2 \varepsilon_{0} A}{d} k_{4}}$

$\Rightarrow \frac{1}{k}=\frac{3}{2\left(k_{1}+k_{2}+k_{3}\right)}+\frac{1}{2 k_{4}} \therefore \frac{2}{k}=\frac{3}{k_{1}+k_{2}+k_{3}}+\frac{1}{k_{4}}$