Answer is (B)
When capacitors are connected is series $C_{e q}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}$ and the potential areas plates of capacitors is given by $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$

If we enter the dielectric material of constant is in capacitor say $C_{2}$ as shown then, the equivalent capacitance of $C_{2}=k \frac{\varepsilon_{0} A}{d} \Rightarrow$ K.G.
$\therefore \quad C_{e q}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{C_{1} \cdot k C_{1}}{C_{1}+C_{1}}=\frac{k C_{1}}{2}$
$\therefore$ The potential plates of $C_{2}$ will be
$\mathrm{V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}}=\frac{\mathrm{Q}}{\mathrm{kC}_{1}}$
As there is only air between the plates of capacitor $C_{1}$ and $(R)=1$ for air
$\mathrm{V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}$
$\begin{array}{l}
\therefore \quad\left(V_{2}\right)_{\text {eff }}=\frac{Q}{C_{1}}\left(\frac{1}{k}+\frac{f 1}{1}\right)=\left(\frac{k+1}{k}\right) \frac{Q}{C_{1}}=\left(\frac{k_{2}+1}{k}\right) V_{1} \\
\therefore \quad V_{1}=\left(\frac{k}{k+1}\right) V_{2}
\end{array}$
If $\quad V_{2}=V$ then $V_{1}=\frac{K \cdot V}{k+1}$