General approach. For a uniform rod of mass $m$ and length $l$ about an axis through one end making angle $\alpha$ with the rod, $I=\frac{m{l}^2}{3}{\sin }^2\alpha$. For rods whose centre is offset from the axis at perpendicular distance $d$, add $m{d}^2$ using the parallel axis theorem (with the centre-of-mass term $\frac{m{l}^2}{12}{\sin }^2\alpha$ if the rod is not through the axis).

(P) Two rods CA and CB at 90°, axis OCO' bisecting the right angle. Each rod makes 45° with the axis and passes through C, which lies on the axis. So each contributes $\frac{m{l}^2}{3}{\sin }^{2}45^{\circ }=\frac{m{l}^2}{6}$. Total $I_P=2\times \frac{m{l}^2}{6}=\frac{m{l}^2}{3}$. P $\to$ 5.

(Q) Equilateral triangle, axis through vertex C parallel to base AB. The two upper rods CA, CB pass through C and each make 60° with the axis: contribution $2\times \frac{m{l}^2}{3}{\sin }^{2}60^{\circ }=2\times \frac{m{l}^2}{3}\cdot \frac{3}{4}=\frac{m{l}^2}{2}$. The base AB lies parallel to the axis at perpendicular distance $\frac{l\sqrt{3}}{2}$ from it, contributing $0+m{\left(\frac{l\sqrt{3}}{2}\right)}^2=\frac{3m{l}^2}{4}$ (rod parallel to axis has no ${\sin }^2\alpha$ term; only the parallel-axis offset matters). Total $I_Q=\frac{m{l}^2}{2}+\frac{3m{l}^2}{4}=\frac{5m{l}^2}{4}$. Q $\to$ 1.

(R) Rhombus with 90° vertices, axis along diagonal CA. All four rods pass through either C or A (both on the axis) and each makes 45° with the axis. Each contributes $\frac{m{l}^2}{3}{\sin }^{2}45^{\circ }=\frac{m{l}^2}{6}$. Total $I_R=4\times \frac{m{l}^2}{6}=\frac{2m{l}^2}{3}$. R $\to$ 4.

(S) Two rods CA, CB joined at C with axis through C at 30° to each rod. Each rod passes through C and makes 30° with the axis: contribution $2\times \frac{m{l}^2}{3}{\sin }^{2}30^{\circ }=2\times \frac{m{l}^2}{3}\cdot \frac{1}{4}=\frac{m{l}^2}{6}$. Total $I_S=\frac{m{l}^2}{6}$. S $\to$ 2.

Mapping P $\to$ 5, Q $\to$ 1, R $\to$ 4, S $\to$ 2 corresponds to option (A).