Step 1: Identify the most reactive proton in the system, not just the mechanism steps.
A Grignard reagent, $R-MgBr$, behaves as a very strong base because the carbon attached to magnesium carries a partial negative charge. Whenever a Grignard reagent is placed near any compound with an acidic $O-H$ or $N-H$ proton, it reacts with that proton first, faster than almost any other reaction it could do.
Step 2: Recognize methanol as the acidic partner here.
$CH_3OH$ has an $O-H$ proton that, while weakly acidic compared to a carboxylic acid, is still far more acidic than any $C-H$ bond nearby. This makes it the proton the Grignard reagent will grab.
Step 3: Write this as a simple acid-base reaction.
\[ R-MgBr + CH_3-O-H \to R-H + CH_3-O-MgBr \]
The Grignard carbon picks up the acidic hydrogen from the alcohol, and the alkoxide takes the magnesium bromide part.
Step 4: Note what this means for the product.
Because the reaction is really just proton transfer to the carbanion-like carbon, the organic product is simply the alkane $R-H$, with an extra hydrogen replacing the magnesium bromide, not any oxygen containing product.
Step 5: Rule out the other options.
Any option showing an alcohol, ether, or carbon-carbon coupled product would require a different kind of reaction (such as addition to a carbonyl), which is not what is happening with a simple alcohol as the reacting partner.
