B

Step 1: When the capacitors are connected in parallel and charged, their equivalent capacitance, \(C_{eq}\), is given by \(C_{eq} = C + 2C = 3C\).

Step 2: The total charge \(Q\) stored in the system is \(Q = C_{eq} \cdot V = 3C \cdot V\).

Step 3: After disconnecting the battery and filling the dielectric in capacitor C, the new capacitance of capacitor C becomes \(C' = K imes C\).

Step 4: The total capacitance of the capacitors in parallel then becomes \(C_{new} = K imes C + 2C = (K + 2)C\).

Step 5: Since charge remains constant, we have \(Q = C_{new} \cdot V_{new}\) where \(V_{new}\) is the new potential difference across the capacitors.

Step 6: Thus, we can express \(V_{new}\) as \(V_{new} = \frac{Q}{C_{new}} = \frac{3CV}{(K + 2)C} = \frac{3V}{K + 2}\).

Step 7: Therefore, the potential difference across the capacitors now becomes \(\frac{3V}{K + 2}\).
Hence, the correct answer is Option B.