Interpret the circuit

This is the standard capacitor-sharing setup:

• Initially, with switch at A, one capacitor gets charged.
• Then switch is moved to B, and the charged capacitor is connected with another capacitor, so charge redistributes.
• The loss in electrostatic energy appears as heat dissipated in the circuit.

Initial state

Let the initially charged capacitor have capacitance $C$ and charge $Q$.

Its initial energy is $U_i=\frac{Q^2}{2C}.$

After switching to B

From the given options and standard configuration, the charged capacitor $C$ is connected to an uncharged capacitor of capacitance $3C$ in parallel.

Hence total capacitance after connection is $C_{\text{eq}}=C+3C=4C.$

Total charge is conserved: $Q_{\text{total}}=Q.$

Therefore final common potential is $V_f=\frac{Q}{4C}.$

Final energy

The final electrostatic energy stored in the two-capacitor combination is $U_f=\frac{Q^2}{2C_{\text{eq}}}=\frac{Q^2}{2(4C)}=\frac{Q^2}{8C}.$

Energy dissipated

$U_{\text{diss}}=U_i-U_f$ $=\frac{Q^2}{2C}-\frac{Q^2}{8C}$ $=\frac{4Q^2-Q^2}{8C}$ $=\frac{3Q^2}{8C}.$

Match with options

$\boxed{\frac{3}{8}\frac{Q^2}{C}}$

So the correct option is D.