Step 1: Recall Cannizzaro’s reaction.
Aldehydes without $\alpha$-hydrogens undergo Cannizzaro’s reaction in the presence of strong base. In this reaction, one molecule of aldehyde is reduced to alcohol and another molecule is oxidized to carboxylic acid.
Step 2: Apply to benzaldehyde.
Benzaldehyde has no $\alpha$-hydrogen. So, it undergoes Cannizzaro’s reaction:
\[
2C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa
\]
Here, one molecule forms benzyl alcohol (C$_6$H$_5$CH$_2$OH), and the other forms sodium benzoate.
Step 3: Analyze options.
- (A) Aldol condensation: Requires $\alpha$-hydrogen, not possible in benzaldehyde.
- (B) Cannizzaro’s reaction: Correct, gives benzyl alcohol.
- (C) Gattermann-Koch reaction: Used for formylation of benzene, not applicable here.
- (D) Clemmensen reduction: Reduces carbonyl group to methylene, not giving benzyl alcohol.
Step 4: Conclusion.
Thus, benzyl alcohol is obtained from benzaldehyde by Cannizzaro’s reaction.
Final Answer:
\[
\boxed{\text{(B) Cannizzaro’s reaction}}
\]