A slab of dielectric constant K has the same crosssectional area as...

From the above diagram, we can write,

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

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

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

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

${C}_{1} = \dfrac{{ϵ}_{0}A}{x}$, ${C}_{2} = \dfrac{4K{ϵ}_{0}A}{3d}$ and ${C}_{3} = \dfrac{{ϵ}_{0}A}{y}$

The equivalent capacitance,

$\dfrac{1}{C} = \dfrac{1}{{C}_{1}} + \dfrac{1}{{C}_{2}} + \dfrac{1}{{C}_{3}} ⇒ \dfrac{1}{C} = \dfrac{1}{{ϵ}_{0}A}\left(x + \dfrac{3d}{4K} + y\right) ⇒ \dfrac{1}{C} = \dfrac{1}{{ϵ}_{0}A}\left(\dfrac{d}{4} + \dfrac{3d}{4K}\right) ⇒ C = \dfrac{4{ϵ}_{0}AK}{d\left(3 + K\right)} ⇒ C = \dfrac{4K{C}_{0}}{3 + K}$