The reaction described involves a mixture of p-methoxy benzaldehyde (paramethoxybenzaldehyde) and methanal (formaldehyde) with concentrated caustic soda (sodium hydroxide, NaOH ) solution. In this scenario, the reaction is a crossed Cannizzaro reaction. The Cannizzaro reaction is a redox reaction where an aldehyde is transformed into an alcohol and a carboxylic acid (or its salt) when treated with a strong base. In the crossed Cannizzaro reaction, two different aldehydes that lack alpha-hydrogens (like methanal and p-methoxy benzaldehyde) react. The more electrophilic aldehyde (typically the one more capable of stabilizing negative charge) will tend to form the carboxylate, while the other aldehyde gets reduced to the corresponding alcohol. p-Methoxy benzaldehyde has the ability to stabilize negative charge through resonance due to the electron-donating methoxy group which makes the aldehyde carbon less electrophilic compared to methanal. This allows methanal to preferentially form the carboxylate salt under the reaction conditions. Therefore, in the presence of a strong base like NaOH : Methanal gets oxidized to form sodium formate (HCOO−Na+). p-Methoxy benzaldehyde gets reduced to form p-methoxy benzyl alcohol ( CH3O−C6H4CH2‌OH). Therefore, the correct product of this reaction would be p-Methoxy benzyl alcohol and sodium formate. Looking at the given options, the correct answer is: Option B: p-Methoxy benzyl alcohol and sodium formate.