The correct option is A
$$C=\cfrac QV$$ $$\therefore V=\cfrac QC$$
Whenever capacitors are connected in series, the charge accumulated in their plates is common. Hence $$Q$$ is common.
Initially, the capacitance of both is the same. $$\therefore V$$is same for both.
Whenever a dielectric material is inserted in between the plates of a capacitor, its capacitance increases. $$Now \therefore C_2>C_1$$
If we look at the above equations, we can note that for $$Q$$ to remain common for both, and for $$C_2$$ to increase, $$V_2$$ has to decrease.
$$V= V_1 +V_2$$ here $$V$$ is the PD of the cell.
Since that doesn't change, and if $$V_2$$ decreases, $$V_1$$ has to increase.
The potential difference across $$C_1$$ will increase.