1-Methylcyclohexene is treated with BH3· THF followed by H2O2/NaOH to give P. P is oxidised with pyridinium chlorochromate to Q, and Q is then…

Hydroboration-oxidation adds $\mathrm{H}$ and $\mathrm{OH}$ across the double bond in an anti-Markovnikov, syn manner, so the boron and hence the hydroxyl group end up on the less substituted carbon $\mathrm{C2}$. $\mathbf{P}$ is therefore trans-2-methylcyclohexanol, a secondary alcohol, and not the tertiary 1-methylcyclohexanol. PCC oxidises a secondary alcohol cleanly to the ketone, so $\mathbf{Q}$ is 2-methylcyclohexanone, $\mathrm{C_7H_{12}O}$. Under base each enolisable $\alpha$-hydrogen is exchanged for deuterium through the enolate. The $\alpha$-carbons are $\mathrm{C2}$ (one H, bearing the methyl) and $\mathrm{C6}$ (two H); the methyl hydrogens are not $\alpha$ to the carbonyl. Three hydrogens are exchanged, giving $\mathrm{C_7H_9D_3O}$. Option (D) would be the answer if the tertiary alcohol were formed (no oxidation, no exchange), and (B) or (C) miscount the $\alpha$-hydrogens.