In the following sequence of reactions:

The correct answer is n-propyl alcohol i.e. option C which can be explained as follows:

Mixture of P + I₂ on reaction with ethanol causes its iodination i.e.  product A is ethyl iodide(CH₃CH₂I) which on reaction with Mg, ether gives grignard reagent, CH₃CH₂MgI (product B) which reacts with formaldehyde to form CH₃CH₂CH₂-OMgI (product C) which finally on hydrolysis gives n-propyl alcohol.

Hence, the correct answer is option C,  CH₃CH₂CH₂-OH.