An ideal 12 V battery is connected through a switch to a resistor R1=4 Ω in series with a parallel combination of an ideal inductor L=15 mH and a…

Just after closing, an inductor with no initial current behaves as an open circuit, so the current flows through $R_1$ and $R_2$ in series: $I=\dfrac{12}{4+12}=0.75\ \mathrm{A}$, matching (P)$\to$3. The voltage across the parallel branch, and so across the inductor, is then $12-0.75\times4=9\ \mathrm{V}$, so (R)$\to$1 (not $12\ \mathrm{V}$, which would ignore the drop across $R_1$). A long time later the inductor current is steady and the ideal inductor acts as a short circuit, bypassing $R_2$, so the battery current is $12/4=3\ \mathrm{A}$, giving (Q)$\to$4. With the ideal battery replaced by a short, the inductor sees $R_1$ in parallel with $R_2$ as its Thevenin resistance, i.e. $\dfrac{4\times12}{4+12}=3\ \mathrm{\Omega}$, so $\tau=L/R_{th}=15\ \mathrm{mH}/3\ \Omega=5\ \mathrm{ms}$, giving (S)$\to$2. Hence P$\to$3, Q$\to$4, R$\to$1, S$\to$2.