$\mathrm{R}-\mathrm{C}\equiv \mathrm{N}\stackrel{{\mathrm{LiAlH}}_4}{\to }\mathrm{Q}$

$\mathrm{R}-\mathrm{C}\equiv \mathrm{N}\stackrel{\mathrm{RMgX}+{\mathrm{H}}_2\mathrm{O}/{\mathrm{H}}^{+}}{\to }\mathrm{P}$

the products $P$ and $Q$ are respectively

The products P and Q are Ketone and 1 amine respectively. 

• Alkyl cyanide reacts with Grignard reagent (followed by acid hydrolysis) to form a ketone. 
• Alkyl cyanide on reduction with lithium aluminum hydride forms primary amine. Following is the reaction that takes place;

               $R-C\equiv N+{LiAlH}_4\to R-{CH}_2-{NH}_2\left(amine\right)$

Hence, the correct answer is B) ketone and $1°$ - amine