Two condensers of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K$. The p.d. across the capacitors will now be

Initially, the equivalent capacitance is $2C + C = 3C$. So the total charge stored before removing the battery is $Q = 3CV$. After the battery is removed, this total charge remains constant because the parallel combination is isolated. When the capacitor of capacity $C$ is completely filled with dielectric constant $K$, its new capacitance becomes $KC$. Hence the new equivalent capacitance is $2C + KC = C(K+2)$. Therefore the new potential difference is $$V' = \frac{Q}{C(K+2)}.$$ Substituting $Q = 3CV$, we get $$V' = \frac{3CV}{C(K+2)} = \frac{3V}{K+2}.$$ This matches option $A$.