The major product (P) in the following reaction is:

The starting material is phenylglyoxal, $$\text{Ph-C(=O)-CHO},$$ which is an $\alpha$-ketoaldehyde. It contains two carbonyl groups but does not possess any $\alpha$-hydrogen atoms.

The reaction is carried out in the presence of $$KOH$$ and heat $$\Delta.$$

When an aldehyde lacking $\alpha$-hydrogens is treated with a strong base, it undergoes the Cannizzaro reaction. Since phenylglyoxal contains both an aldehyde and a ketone group within the same molecule, it undergoes an intramolecular Cannizzaro reaction.

In a Cannizzaro reaction, one carbonyl group is oxidized while the other is reduced. Here, the aldehyde portion is oxidized to a carboxylate ion, whereas the ketone portion is reduced to an alcohol.

The mechanism proceeds as follows:

1. The hydroxide ion $$OH^-$$ attacks the more electrophilic and less sterically hindered aldehyde carbonyl carbon.
2. The resulting alkoxide intermediate undergoes a hydride shift. The negatively charged oxygen reforms the carbonyl group, causing the adjacent hydrogen atom to migrate as a hydride ion $$H^-$$ to the neighboring ketone carbonyl carbon.
3. This produces a carboxylate group and an alkoxide ion. A rapid proton transfer then converts the alkoxide into a secondary alcohol while the oxidized portion remains as the potassium carboxylate salt.

Thus,

• the aldehyde group $$-CHO$$ is converted into $$-COO^-K^+,$$
• and the ketone group $$-C(=O)-$$ is converted into $$-CH(OH)-.$$

The final product obtained is potassium mandelate, having the structure

$$\boxed{\text{Ph-CH(OH)-COO}^-,\text{K}^+.}$$

Therefore, the correct answer is

$$\boxed{\text{Option (B).}}$$