In the given reaction, the product 'A' is

When an unsymmetrical alkyne reacts with $$\text{H}_2\text{SO}_4$$ and $$\text{HgSO}_4$$, water adds across the triple bond following Markovnikov's rule. The electrophile ($$\text{H}^+$$) attacks the carbon that yields the more stable carbocation intermediate.

The triple bond is between the carbon attached to the phenyl ring ($$\text{C}_1$$) and the carbon attached to the methyl group ($$\text{C}_2$$):

$$\text{Ph}-\text{C}\equiv\text{C}-\text{CH}_3$$

Because the benzylic carbocation is significantly more stable, the nucleophile ($$-\text{OH}$$ from water) will attack $$\text{C}_1$$.

The addition of $$-\text{H}$$ and $$-\text{OH}$$ results in an unstable enol intermediate:

$$\text{Ph}-\overset{\text{OH}}{\text{C}}=\text{CH}-\text{CH}_3$$

Enols are generally unstable and rapidly undergo tautomerization to form the more thermodynamically stable carbonyl compound (keto form).

$$\text{Ph}-\overset{\text{OH}}{\text{C}}=\text{CH}-\text{CH}_3 \rightleftharpoons \text{Ph}-\overset{\text{O}}{\overset{\parallel}{\text{C}}}-\text{CH}_2-\text{CH}_3$$