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Correct option is D)

The common mistake students do in this problem is that they assume $C_{1}$ and $C_{2}$ in parallel and their combination in series with $C_{3}$. Which is completely wrong method. [Refer Fig. 1]

For two capacitors to be taken in parallel, their both ends should be at same potential.

Here, the upper ends of $C_{1}$ and $C_{2}$ are at same potential as they are connected to A through a metal plate.

But their lower ends being connected to a dielectric $K_{3}$ are not at same potential. As Electric field will be different in both dielectrics $K_{1}$ and $K_{2}$, since $K_{1}$ $=$ $K_{2}$.

$Step 1: DivideK_{3}region into two equal parts and find capacitance of each parts [Refer Fig. 2]$

Hence, To solve this problem, we divide dielectric $K_{3}$ in two equal parts as shown in figure 2. So,

$C_{1}=d/2K_{1}ϵ_{0}(A/2) =dK_{1}ϵ_{0}A $

$C_{2}=d/2K_{2}(ϵ_{0})(A/2) =dK_{2}ϵ_{0}A $

$C_{3}=d/2K_{3}ϵ_{0}(A/2) =dK_{3}(ϵ_{0}A) $

$C_{4}=d/2K_{3}ϵ_{0}(A/2) =dK_{3}ϵ_{0}A $

Now, From figure, Clearly, we can assume $C_{1}$ and $C_{3}$ in series and $C_{2}$ and $C_{4}$, further their combinations will be in parallel.

$Step 2: Equivalent capacitance [Refer Fig. 3]$

Combining series combinations in figure 3, we get:

$C_{eq}=C_{1}+C_{3}C_{1}C_{3} +C_{2}+C_{4}C_{2}C_{4} $

Putting values of $C_{1},C_{2},C_{3}$ and $C_{4}$ from Step $(1)$

$C_{eq}=K_{1}+K_{3}K_{1}K_{3} dϵ_{0}A +K_{2}+K_{3}K_{2}K_{3} dϵ_{0}A $

$Step 3: To find$ $K_{eq}$, $CompareC_{eq}withdK_{eq}ϵ_{0}A $ $[Refer Fig. 4]$

$C_{eq}=dK_{eq}ϵ_{0}A $

$⇒dK_{eq}ϵ_{0}A =K_{1}+K_{3}K_{1}K_{3} dϵ_{0}A +K_{2}+K_{3}K_{2}K_{3} dϵ_{0}A $

$⇒$ $K_{eq}=K_{1}+K_{3}K_{1}K_{3} +K_{2}+K_{3}K_{2}K_{3} $

Hence equivalent dielectric constant will be as given in Option $D$

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