Refer image 1.

Let $k(x)=k(1+\alpha x)$

be the dielectric constant

The capacitor can be considered to be a series combination of a large number of capacitors of thickness $\Delta x$.

Refer image 2.

$\frac{1}{C_{eq}}=\sum \frac{\Delta x}{k(x)A{ϵ}_0}={\int }_0^d\frac{dx}{k(x)A{ϵ}_0}$

where $\Delta x\to 0$

$\frac{1}{C_{eq}}={\int }_0^d\frac{dx}{k(1+\alpha x)A{ϵ}_0}=\frac{1}{KA{ϵ}_0}{\int }_0^d\frac{dx}{1+\alpha x}$

 

$=\frac{1}{KA{ϵ}_0}{\log }_e(\frac{1+\alpha d}{1})\times \frac{1}{\alpha }$

$=\frac{1}{K\alpha A{ϵ}_0}{\log }_e(1+\alpha d)$

Now, $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-------$

$\therefore {\log }_e(1+\alpha d)\simeq \alpha d-\frac{{\alpha }^2{d}^2}{2}$

$\therefore \frac{1}{C_{eq}}=\frac{1}{k\alpha A{ϵ}_e}{\log }_e(1+\alpha d)$

$=\frac{1}{k\alpha A{ϵ}_0}[\alpha d-\frac{{\alpha }^2{d}^2}{2}+-------]$

$\simeq \frac{\alpha d}{k\alpha A{ϵ}_0}[1-\frac{\alpha d}{2}]\simeq \frac{d}{KA{ϵ}_0}[1-\frac{\alpha d}{2}]$

$C_{eq}=\frac{KA{ϵ}_0}{d[1-\frac{\alpha d}{2}]}=\frac{KA{ϵ}_0}{d}[1-\frac{\alpha d}{2}{]}^{-1}$

Now, $(1-x{)}^{-1}\simeq 1+x$   $[\because x<<1]$

$\therefore C_{eq}=\frac{KA{ϵ}_0}{d}[1+\frac{\alpha d}{2}]$

option (C) is correct.