The correct answer is $\boxed{\text{(A)}}$.
In the reaction, $\ce{Ph3P ->[Mel] [X] ->[n-BuLi] [Y]}$, the compound $\ce{X}$ is a triphenylphosphonium iodide, $\ce{[Ph3P(Me)I]}$. This is formed when the triphenylphosphine, $\ce{Ph3P}$, reacts with methyl iodide, $\ce{MeI}$, in the presence of a Lewis acid, such as $\ce{Mel}$. The reaction is a nucleophilic substitution reaction, in which the iodide ion, $\ce{I-}$, attacks the phosphorus atom in the triphenylphosphine, displacing the methyl group, $\ce{Me}$.
The compound $\ce{Y}$ is a triphenylphosphine methylene, $\ce{[Ph3P(Me)] [I]}$. This is formed when the triphenylphosphonium iodide, $\ce{[Ph3P(Me)I]}$, reacts with n-butyllithium, $\ce{n-BuLi}$. The reaction is a lithium-halogen exchange reaction, in which the lithium ion, $\ce{Li+}$, attacks the iodide ion, $\ce{I-}$, displacing the methyl group, $\ce{Me}$.
The other options are incorrect. Option $\text{(B)}$ is incorrect because it does not have a methylene group, $\ce{CH2}$. Option $\text{(C)}$ is incorrect because it does not have an iodide ion, $\ce{I-}$. Option $\text{(D)}$ is incorrect because it does not have a triphenylphosphonium ion, $\ce{[Ph3P(Me)I]}$.