Given below are two statements:
Statement I : The nucleophilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.
Statement II : The nucleophilic addition of hydrogen cyanide to an aldehyde or a ketone yields amine as final product.
In the light of the above statements, choose the most appropriate answer from the options given below:

For any carbonyl compound, that is, an aldehyde $$\mathrm{R{-}CHO}$$ or a ketone $$\mathrm{R_2C{=}O}$$, the carbon of the carbonyl group carries a partial positive charge because the double-bonded oxygen is more electronegative and withdraws electron density. Hence this carbon is susceptible to attack by nucleophiles.

First we examine Statement I. In the sodium hydrogen sulphite reaction the actual nucleophile is the bisulphite ion $$\mathrm{HSO_3^-}$$. According to the general mechanism of nucleophilic addition to a carbonyl group, the very first step is:

$$$\mathrm{R_2C{=}O + HSO_3^- \;\longrightarrow\; R_2C(OH)(SO_3H)^{-}}$$$

Here the $$\mathrm{HSO_3^-}$$ ion donates its lone pair to the electrophilic carbon, the $$\pi$$ bond shifts onto oxygen and we obtain an alkoxide ion $$\mathrm{O^-}$$ attached to $$\mathrm{SO_3H}$$. Because we have generated a basic (negatively charged) oxygen and at the same time possess an acidic hydrogen on the $$\mathrm{SO_3H}$$ group, an intramolecular proton transfer immediately follows:

$$$\mathrm{R_2C(} \underset{\small{-}}{O}\mathrm{)(SO_3H) \;+\; H\,(SO_3) \;\longrightarrow\; R_2C(OH)(SO_3H)}$$$

This proton transfer neutralises the alkoxide and produces a stable bisulphite addition product, which is often isolated as a crystalline solid. Because the success of the reaction relies precisely on this proton shift that converts an unstable alkoxide into a neutral alcohol function, the description given in Statement I is accurate. Hence, Statement I is true.

Now we turn to Statement II. The classical nucleophilic addition of hydrogen cyanide to a carbonyl compound is described by the following sequence. First, in the presence of a weak base such as $$\mathrm{NaCN}$$, a small amount of $$\mathrm{HCN}$$ dissociates to give the actual nucleophile $$\mathrm{CN^-}$$:

$$\mathrm{HCN \;\rightleftharpoons\; H^+ + CN^-}$$

The cyanide ion then attacks the carbonyl carbon exactly as any nucleophile would:

$$$\mathrm{R_2C{=}O + CN^- \;\longrightarrow\; R_2C(} \underset{\small{-}}{O}\mathrm{)CN}$$$

The resulting alkoxide immediately abstracts a proton from $$\mathrm{HCN}$$ (or from any other proton donor in the medium) to give what is known as a cyanohydrin:

$$$\mathrm{R_2C(} \underset{\small{-}}{O}\mathrm{)CN + HCN \;\longrightarrow\; R_2C(OH)CN + CN^-}$$$

Thus the direct product of the reaction is $$\mathrm{R_2C(OH)CN}$$, which contains both a hydroxyl (-OH) and a nitrile (-CN) group. No amine (-NH2) function is produced in this step. An amine could appear only after a subsequent hydrogenation of the nitrile group, but that is a completely separate reaction. Therefore Statement II, which claims that the nucleophilic addition of hydrogen cyanide yields amine as the final product, is false.

Summarising, Statement I is true while Statement II is false. Among the given options, this corresponds to Option D.

Hence, the correct answer is Option D.