A parallel plate capacitor of area A, plate separation d and...

The given capacitor can be considered as a combination of four smaller capacitors.

The top half consists of three capacitors in parallel, each with area $A/3$ and thickness $d/2$. Their capacitances are:
$C_1 = \dfrac{k_1 \epsilon_0 (A/3)}{d/2} = \dfrac{2k_1 \epsilon_0 A}{3d}$
$C_2 = \dfrac{k_2 \epsilon_0 (A/3)}{d/2} = \dfrac{2k_2 \epsilon_0 A}{3d}$
$C_3 = \dfrac{k_3 \epsilon_0 (A/3)}{d/2} = \dfrac{2k_3 \epsilon_0 A}{3d}$

The equivalent capacitance of this top part is:
$C_{top} = C_1 + C_2 + C_3 = \dfrac{2\epsilon_0 A}{3d}(k_1 + k_2 + k_3)$

The bottom half is a single capacitor with area $A$ and thickness $d/2$. Its capacitance is:
$C_4 = \dfrac{k_4 \epsilon_0 A}{d/2} = \dfrac{2k_4 \epsilon_0 A}{d}$

The top part $C_{top}$ and the bottom part $C_4$ are in series. The equivalent capacitance $C$ of the entire system is given by:
$\dfrac{1}{C} =\dfrac{1}{C_{top}} + \dfrac{1}{C_4} <br>\dfrac{1}{C} = \dfrac{3d}{2\epsilon_0 A (k_1 + k_2 + k_3)} + \dfrac{d}{2k_4 \epsilon_0 A}$

If a single dielectric of constant $k$ is used, the capacitance would be:
$C = \dfrac{k \epsilon_0 A}{d} \Rightarrow \dfrac{1}{C} = \dfrac{d}{k \epsilon_0 A}$

Equating the two expressions for $\dfrac{1}{C}$ :
$\dfrac{d}{k \epsilon_0 A} = \dfrac{d}{2\epsilon_0 A} \left( \dfrac{3}{k_1 + k_2 + k_3} + \dfrac{1}{k_4} \right)$

Canceling $\dfrac{d}{\epsilon_0 A}$ from both sides and multiplying by 2:
$\dfrac{2}{k} = \dfrac{3}{k_1 + k_2 + k_3} + \dfrac{1}{k_4} <br><br>Answer: \dfrac{2}{k} = \dfrac{3}{k_1 + k_2 + k_3} + \dfrac{1}{k_4}$