The major product formed in the reaction is

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(A) The starting material is a $\gamma$-lactone (a cyclic ester). When a lactone reacts with an excess of a Grignard reagent $(PhMgBr)$,the reaction p

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(A) The starting material is a $\gamma$-lactone (a cyclic ester). When a lactone reacts with an excess of a Grignard reagent $(PhMgBr)$,the reaction proceeds in two steps: $1$. The Grignard reagent attacks the carbonyl carbon of the ester,leading to the opening of the ring and the formation of a ketone intermediate. $2$. Because the Grignard reagent is in excess,it immediately attacks the newly formed ketone to produce a tertiary alcohol. Specifically,the ring opens at the ester linkage,and two phenyl groups are added to the original carbonyl carbon,while the oxygen of the carbonyl becomes a hydroxyl group. The original ether oxygen of the lactone becomes a primary or secondary alcohol group depending on the structure. In this case,the $\gamma$-lactone with a methyl group at the $\alpha$-position reacts with $PhMgBr$ to yield a diol where the tertiary alcohol is formed at the carbon that was originally the carbonyl carbon,and the secondary alcohol is formed at the carbon that was originally attached to the ring oxygen. The correct structure is shown in option $A$.

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