In the reaction, 

${\mathrm{CH}}_3·{\mathrm{CH}}_2·\mathrm{Cl}\stackrel{\mathrm{KCN}}{⟶}\left(\mathrm{A}\right)\stackrel{{\mathrm{LiAlH}}_4}{⟶}\left(\mathrm{P}\right)$ end product $\left(\mathrm{P}\right)$ is

When the reducing agent is lithium aluminium hydride, the aldehyde group is reduced to primary alcohol but the carbon-carbon double bond is unaffected.

In the given reaction the end product is (d). It is obtained
as follows.

${\mathrm{CH}}_3{\mathrm{CH}}_2·\mathrm{Cl}\stackrel{\mathrm{KCN}}{⟶}\underset{\left(\mathrm{A}\right)}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CN}}\underset{\left(\mathrm{Reduction}\right)}{\overset{{\mathrm{LiAlH}}_4}{\to }}{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2·{\mathrm{NH}}_2$