But-2-yne (X) is converted in two ways. Route 1: (i) Na/liq. NH3; (ii) Br2/CCl4; (iii) one equivalent of alcoholic KOH (E2, anti-periplanar…

Route 1: dissolving-metal reduction gives trans-but-2-ene; anti addition of $\mathrm{Br_2}$ to a trans alkene gives meso-2,3-dibromobutane ($2R,3S$). Route 2: Lindlar reduction gives cis-but-2-ene; anti addition of bromine gives the racemic ($2R,3R$/$2S,3S$) dibromide. For E2 the $\mathrm{H}$ and $\mathrm{Br}$ removed must be anti-periplanar. Placing the meso dibromide in the conformation with $\mathrm{Br}$ (on C2) anti to $\mathrm{H}$ (on C3) puts the two methyl groups gauche, so they end up cis on the new double bond. On C3 the higher priority group is $\mathrm{Br}$ and on C2 it is $\mathrm{CH_3}$; with the methyls cis these lie on opposite sides, which is the $E$ isomer. The racemic dibromide, by the same analysis, has its methyls anti and gives the $Z$ isomer. So P is $E$ and Q is $Z$, option B.