Phenyl magnesium bromide reacts with methanol to give

Phenyl magnesium bromide is a Grignard reagent, written as $$Ph{-}MgBr$$, where the carbon bearing the phenyl group behaves as a strong carbanion ($$Ph^-$$) and, therefore, as a very strong base.

Rule for Grignard reagents: they react immediately with any proton source $$\bigl(R{-}OH,\; H_2O,\; RCO_2H,\; NH_3,\ldots\bigr)$$ to abstract a proton and give the corresponding hydrocarbon plus an alkoxide (or related) magnesium halide.

Methanol is a protic solvent and supplies an acidic proton from its -OH group:
$$MeOH \; \longrightarrow \; MeO^- + H^+$$ (conceptually).

Therefore, phenyl magnesium bromide reacts by simple acid-base neutralisation:

$$Ph{-}MgBr \;+\; MeOH \;\longrightarrow\; Ph{-}H \;+\; MeO{-}MgBr$$

Identifying the products:
• $$Ph{-}H$$ is benzene.
• $$MeO{-}MgBr$$ is magnesium bromomethoxide, usually written as $$Mg(OMe)Br$$.

Hence, the reaction yields a mixture of benzene and $$Mg(OMe)Br$$.

So, the correct choice is:
Option B which is: a mixture of benzene and Mg(OMe)Br