To solve this, we analyze the arrangement of dielectrics based on the resulting formula (standard NEET 2016 configuration):

1. Geometry Analysis: The capacitor is split into two halves along the plate separation $d$.

• Top Half (thickness $d/2$): Divided into three equal areas ($A/3$) filled with $k_1$, $k_2$, and $k_3$. These form three capacitors in parallel.
• Bottom Half (thickness $d/2$): Filled completely with $k_4$ (Area $A$). This forms a capacitor in series with the top combination.

2. Top Capacitance ($C_{top}$): The capacitance of a parallel plate capacitor is $C = \frac{K \varepsilon_0 A}{d}$. For the top three parts: $C_1 = \frac{k_1 \varepsilon_0 (A/3)}{d/2} = \frac{2 k_1 \varepsilon_0 A}{3d}$, $C_2 = \frac{2 k_2 \varepsilon_0 A}{3d}$, $C_3 = \frac{2 k_3 \varepsilon_0 A}{3d}$. In parallel, $C_{top} = C_1 + C_2 + C_3 = \frac{2 \varepsilon_0 A}{3d} (k_1 + k_2 + k_3)$.

3. Bottom Capacitance ($C_{bot}$): $C_4 = \frac{k_4 \varepsilon_0 A}{d/2} = \frac{2 k_4 \varepsilon_0 A}{d}$.

4. Equivalent Capacitance ($C_{eq}$): The top and bottom systems are in series. $\frac{1}{C_{eq}} = \frac{1}{C_{top}} + \frac{1}{C_{bot}}$. We want the equivalent dielectric $k$ such that $C_{eq} = \frac{k \varepsilon_0 A}{d}$. Substituting: $\frac{d}{k \varepsilon_0 A} = \frac{1}{\frac{2 \varepsilon_0 A}{3d} (k_1 + k_2 + k_3)} + \frac{1}{\frac{2 k_4 \varepsilon_0 A}{d}}$ $\frac{d}{k \varepsilon_0 A} = \frac{3d}{2 \varepsilon_0 A (k_1 + k_2 + k_3)} + \frac{d}{2 k_4 \varepsilon_0 A}$

5. Simplify: Divide by $\frac{d}{\varepsilon_0 A}$: $\frac{1}{k} = \frac{3}{2(k_1 + k_2 + k_3)} + \frac{1}{2k_4}$ Multiply by 2: $\frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}$