Total deviation from the standard geometry of refraction-reflection-refraction inside a sphere:

$\alpha =({\theta }_0-{\varphi }_0)+(\pi -2{\varphi }_0)+({\theta }_0-{\varphi }_0)=\pi +2{\theta }_0-4{\varphi }_0$.

Snell at entry: $\sin {\theta }_0=n\sin {\varphi }_0$.

(P) $n=\sqrt{2}$, $\alpha =180^{\circ }=\pi$. Then $2{\theta }_0=4{\varphi }_0$, so ${\theta }_0=2{\varphi }_0$. Substituting in Snell: $\sin (2{\varphi }_0)=\sqrt{2}\sin {\varphi }_0\Rightarrow 2\sin {\varphi }_0\cos {\varphi }_0=\sqrt{2}\sin {\varphi }_0$. Solutions: $\sin {\varphi }_0=0$ (${\varphi }_0=0$, hence ${\theta }_0=0$) or $\cos {\varphi }_0=1/\sqrt{2}$ (${\varphi }_0=45^{\circ }$, ${\theta }_0=90^{\circ }$). The non-trivial ${\theta }_0=90^{\circ }$ does not fit in the options. The trivial solution ${\theta }_0=0^{\circ }$ matches option (5). So P $\to$ 5.

(Q) $n=\sqrt{3}$, same condition ${\theta }_0=2{\varphi }_0$. $\sin (2{\varphi }_0)=\sqrt{3}\sin {\varphi }_0\Rightarrow \cos {\varphi }_0=\sqrt{3}/2$, so ${\varphi }_0=30^{\circ }$, ${\theta }_0=60^{\circ }$ (or trivial $0^{\circ }$). ${\theta }_0$ values: $60^{\circ }$ and $0^{\circ }$. Q $\to$ 2.

(R) Same case, but listing ${\varphi }_0$ values: $30^{\circ }$ and $0^{\circ }$. R $\to$ 1.

(S) $n=\sqrt{2}$, ${\theta }_0=45^{\circ }$. Snell: $\sin 45^{\circ }=\sqrt{2}\sin {\varphi }_0\Rightarrow \sin {\varphi }_0=1/2\Rightarrow {\varphi }_0=30^{\circ }$. $\alpha =180^{\circ }+2(45^{\circ })-4(30^{\circ })=180^{\circ }+90^{\circ }-120^{\circ }=150^{\circ }$. S $\to$ 4.

$\therefore$ Option (A) is correct.