A slab of dielectric constant K has the same crosssectional area as the plates of a parallel plate capacitor…

From the above diagram, we can write,

$x+y+\frac{3d}{4}=d$

$x+y=\frac{d}{4}$

Initially, when there is no dielectric inserted between the plates the capacitance will be, $C_0=\frac{{ϵ}_{0}A}{ d}$.

When the dielectric slab is inserted, we can assume there are $3$ capacitors that are connected in series.

$C_1=\frac{{ϵ}_{0}A}{x}$, $C_2=\frac{4K{ϵ}_{0}A}{3d}$ and $C_3=\frac{{ϵ}_{0}A}{y}$

The equivalent capacitance,

$$\begin{gathered} \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \\ \Rightarrow \frac{1}{C}=\frac{1}{{ϵ}_{0}A}\left(x+\frac{3d}{4K}+y\right) \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle C}}=\frac{{\displaystyle 1}}{{\displaystyle {ϵ}_{0}A}}\left(\frac{d}{4}+\frac{3d}{4K}\right) \\ \Rightarrow C=\frac{4{ϵ}_{0}AK}{d\left(3+K\right)} \\ \Rightarrow C=\frac{4K{C}_0}{3+K} \end{gathered}$$