Benzaldehyde and an excess of formaldehyde are treated with concentrated aqueous NaOH (crossed Cannizzaro reaction).

Neither aldehyde has an $\alpha$-hydrogen, so hydroxide adds to a carbonyl to give a hydrate anion that transfers hydride to a second aldehyde molecule. Formaldehyde is far more electrophilic and is hydrated much more easily, so its hydrate anion is the hydride donor and it is oxidised to formate, while benzaldehyde, the hydride acceptor, is reduced to benzyl alcohol; (A) and (B) are correct and (C) is wrong because the donor is formaldehyde, not benzaldehyde. With $\mathrm{CD_2O}$ the transferred species is a deuteride, which ends on the carbonyl carbon of benzaldehyde: $\mathrm{C_6H_5CHO + D^-\to C_6H_5CHD\text{-}O^-}$, giving $\mathrm{C_6H_5CHDOH}$ (racemic, with the formate being $\mathrm{DCOO^-}$); (D) is correct.