net capacitance of the system is series combination of "elemental capacitors" each of thickness $dx$

$\dfrac{1}{C_{net}}=\dfrac{1}{A\epsilon_{0}}\displaystyle \int_{0}^{d}\dfrac{dx}{e^{ax}}\implies C_{net}=\dfrac{aA\epsilon_{0}}{1-e^{-ad}}$

$V=\dfrac{Q}{C_{net}}=Q\left[\dfrac{1-e^{-ad}}{aA\epsilon\_{0}}\right]\tag{1}$

by inserting inserting dielectric D remains same

we have ,

$\vec {D}=electric \ displacement=\epsilon_{0}K E \ { {\hat x}}$

again by definition of D i.e, it equal to the amount of **free charge** on one plate divided by the area of the plate $\left(\dfrac{Q}{A}\right)$

$D=\left(\dfrac{Q}{A}\right)=\epsilon_{0}K E \ { }\implies E=E(x)=\dfrac{Q}{A\epsilon_{0}{K}}$

in above, put$ K=e^{ax}$

$E(x)=\dfrac{Q}{A\epsilon_{0}e^{ax}}$

if you calculate $-\displaystyle \int _{0}^{d} E(x) dx $ using above expression of E(x) you will get same voltage as above **eq.(1)**

i think **answer in your book is wrong**

and also, charge $Q(x)$ will not vary as function of $x$because both $C_{net}$ and $V$ are fixed constant quantities