To solve this problem, we need to determine the dimensional formula for each physical quantity in List I and then match it with the corresponding formula in List II.

P. Boltzmann Constant ($k_B$)

1. The Boltzmann constant relates the average kinetic energy ($E$) of particles in a gas to the absolute temperature ($T$). The formula is given by: $E = \frac{3}{2} k_B T$
2. Rearranging for $k_B$, we get (the constant $\frac{3}{2}$ is dimensionless): $k_B = \frac{E}{T}$
3. The dimensional formula for energy ($E$) is $[ML^2T^{-2}]$.
4. The dimensional formula for temperature ($T$) is $[K]$.
5. Substituting these into the equation for $k_B$: $[k_B] = \frac{[E]}{[T]} = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$
6. This matches with option 4 in List II.

Q. Coefficient of viscosity ($\eta$)

1. The coefficient of viscosity is defined by Newton's law of viscosity for the viscous force ($F$) between two layers of a fluid: $F = \eta A \frac{dv}{dx}$ where $A$ is the area and $\frac{dv}{dx}$ is the velocity gradient.
2. Rearranging for $\eta$: $\eta = \frac{F}{A \frac{dv}{dx}}$
3. The dimensional formulas for the quantities on the right are:
   • Force ($F$): $[MLT^{-2}]$
   • Area ($A$): $[L^2]$
   • Velocity gradient ($\frac{dv}{dx}$): $\frac{[LT^{-1}]}{[L]} = [T^{-1}]$
4. Substituting these into the equation for $\eta$: $[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} = [MLT^{-2}L^{-2}T^1] = [ML^{-1}T^{-1}]$
5. This matches with option 2 in List II.

R. Planck Constant ($h$)

1. The Planck constant relates the energy ($E$) of a photon to its frequency ($\nu$). The formula is: $E = h\nu$
2. Rearranging for $h$: $h = \frac{E}{\nu}$
3. The dimensional formulas are:
   • Energy ($E$): $[ML^2T^{-2}]$
   • Frequency ($\nu$): $[T^{-1}]$
4. Substituting these into the equation for $h$: $[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$
5. This matches with option 1 in List II.

S. Thermal conductivity ($k$)

1. The rate of heat flow ($\frac{dQ}{dt}$) through a material is given by Fourier's law of heat conduction: $\frac{dQ}{dt} = k A \frac{dT}{dx}$ where $\frac{dQ}{dt}$ is power, $A$ is the cross-sectional area, and $\frac{dT}{dx}$ is the temperature gradient.
2. Rearranging for $k$: $k = \frac{(\frac{dQ}{dt})}{A (\frac{dT}{dx})}$
3. The dimensional formulas are:
   • Rate of heat flow / Power ($\frac{dQ}{dt}$): $\frac{[Energy]}{[Time]} = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]$
   • Area ($A$): $[L^2]$
   • Temperature gradient ($\frac{dT}{dx}$): $\frac{[K]}{[L]} = [KL^{-1}]$
4. Substituting these into the equation for $k$: $[k] = \frac{[ML^2T^{-3}]}{[L^2][KL^{-1}]} = [ML^2T^{-3}L^{-2}K^{-1}L^1] = [MLT^{-3}K^{-1}]$
5. This matches with option 3 in List II.

Conclusion

Based on the derivations, the correct matching is:

• P $\rightarrow$ 4
• Q $\rightarrow$ 2
• R $\rightarrow$ 1
• S $\rightarrow$ 3

This corresponds to option C.