2,3-Dimethylbutane-2,3-diol is warmed with conc. H2SO4 to give a ketone P. Treating P with excess I2 in aqueous NaOH and then acidifying the aqueous…

Step 1 is the pinacol-pinacolone rearrangement. Protonation and loss of water from one tertiary $\mathrm{OH}$ gives a tertiary carbocation next to a carbon bearing $\mathrm{OH}$ and two methyl groups; a methyl shift then gives a protonated ketone, and deprotonation gives pinacolone, $\mathbf{P}=\mathrm{(CH_3)_3C{-}CO{-}CH_3}$ (3,3-dimethylbutan-2-one). Step 2 is the haloform reaction, which works for a methyl ketone: the $\mathrm{CH_3}$ is tri-iodinated, hydroxide attacks the carbonyl and $\mathrm{{}^-CI_3}$ leaves. This produces the carboxylate of $\mathrm{(CH_3)_3C{-}COOH}$ and yellow iodoform $\mathrm{CHI_3}$. Option A has an extra carbon, C has the wrong skeleton, and B names methyl iodide, which is not the yellow solid.