The products $\mathrm{P}$ and $\mathrm{Q}$ of the following reaction, respectively, are $\mathrm{a} {\mathrm{S}}_8( \mathrm{s})+\mathrm{bNaOH}\left(\mathrm{aq}\right)⟶\mathrm{P}\left(\mathrm{aq}\right)+\mathrm{qQ}\left(\mathrm{aq}\right)+{\mathrm{rH}}_2\mathrm{O}$

We know that ${\mathrm{S}}_8$ has crown-shaped ring structure while $\mathrm{NaOH}$ is a strong alkali and when sulphur reacts with hot aqueous alkali hydroxide it forms sodium sulfide and thiosulphate species.

So, the balanced chemical equation is represented as 

${\mathrm{S}}_8 + 12 \mathrm{NaOH} \to 4 {\mathrm{Na}}_2\mathrm{S} + 2 {\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3 + 6 {\mathrm{H}}_2\mathrm{O}$

Hence, the obtained products are ${\mathrm{Na}}_2\mathrm{S} \& {\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$