The system consists of four identical parallel metal sheets, which form three capacitors between adjacent sheets. Let the capacitance between any two adjacent sheets be $C_0$. Given that the area of each sheet is $a^2$ and the distance between them is $d$, we have $C_0 = \frac{\varepsilon_0 a^2}{d}$. We will analyze each configuration from List-I to find its equivalent capacitance.

Analysis of each configuration:

(P) The capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected.

1. A potential difference is applied between plates $S_1$ and $S_4$. The plates $S_2$ and $S_3$ are floating.
2. This configuration creates three capacitors in series: the one between $S_1$ and $S_2$ ($C_{12}$), the one between $S_2$ and $S_3$ ($C_{23}$), and the one between $S_3$ and $S_4$ ($C_{34}$). Each has a capacitance of $C_0$.
3. The equivalent capacitance ($C_P$) for capacitors in series is given by: $\frac{1}{C_P} = \frac{1}{C_{12}} + \frac{1}{C_{23}} + \frac{1}{C_{34}} = \frac{1}{C_0} + \frac{1}{C_0} + \frac{1}{C_0} = \frac{3}{C_0}$
4. Therefore, $C_P = \frac{C_0}{3}$.
5. This matches with (3) in List-II. So, P → 3.

(Q) The capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$.

1. A potential difference is applied between plates $S_1$ and $S_4$. Plates $S_2$ and $S_3$ are connected by a wire, making them an equipotential surface.
2. The capacitor between $S_2$ and $S_3$ ($C_{23}$) is short-circuited as there is no potential difference across it. It does not store any charge and can be removed from the effective circuit.
3. The system effectively becomes two capacitors, $C_{12}$ and $C_{34}$, connected in series.
4. The equivalent capacitance ($C_Q$) is: $\frac{1}{C_Q} = \frac{1}{C_{12}} + \frac{1}{C_{34}} = \frac{1}{C_0} + \frac{1}{C_0} = \frac{2}{C_0}$
5. Therefore, $C_Q = \frac{C_0}{2}$.
6. This matches with (2) in List-II. So, Q → 2.

(R) The capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$.

1. A potential difference is applied between plates $S_1$ and $S_3$. Plates $S_2$ and $S_4$ are connected by a wire.
2. Let the potential of $S_1$ be $V_A$ and the potential of $S_3$ be $V_B$. The common potential of $S_2$ and $S_4$ is $V_C$.
3. The capacitors are connected as follows:
   • $C_{12}$ is between $S_1$ (potential $V_A$) and $S_2$ (potential $V_C$).
   • $C_{23}$ is between $S_2$ (potential $V_C$) and $S_3$ (potential $V_B$).
   • $C_{34}$ is between $S_3$ (potential $V_B$) and $S_4$ (potential $V_C$).
4. From this, we see that $C_{23}$ and $C_{34}$ are connected in parallel between points C and B. Their combined capacitance is $C_{parallel} = C_{23} + C_{34} = C_0 + C_0 = 2C_0$.
5. This parallel combination is in series with $C_{12}$ (which is between points A and C).
6. The total equivalent capacitance ($C_R$) between A and B is: $\frac{1}{C_R} = \frac{1}{C_{12}} + \frac{1}{C_{parallel}} = \frac{1}{C_0} + \frac{1}{2C_0} = \frac{2+1}{2C_0} = \frac{3}{2C_0}$
7. Therefore, $C_R = \frac{2C_0}{3}$.
8. This matches with (4) in List-II. So, R → 4.

(S) The capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$.

1. A potential difference is applied between $S_1$ and $S_2$.
2. Plates $S_1$ and $S_3$ are connected, forming one terminal (let's call it A). Plates $S_2$ and $S_4$ are connected, forming the other terminal (let's call it B).
3. Let's look at the connections for the three capacitors:
   • $C_{12}$ is between $S_1$ (terminal A) and $S_2$ (terminal B).
   • $C_{23}$ is between $S_2$ (terminal B) and $S_3$ (terminal A).
   • $C_{34}$ is between $S_3$ (terminal A) and $S_4$ (terminal B).
4. All three capacitors, $C_{12}$, $C_{23}$, and $C_{34}$, are connected between the same two terminals A and B. This is a parallel combination.
5. The equivalent capacitance ($C_S$) is the sum of the individual capacitances: $C_S = C_{12} + C_{23} + C_{34} = C_0 + C_0 + C_0 = 3C_0$
6. This matches with (1) in List-II. So, S → 1.

Summary of Matches:

• P → 3
• Q → 2
• R → 4
• S → 1

This corresponds to option C.