The major product of the following reaction is ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CHCO}}_2{\mathrm{CH}}_3\stackrel{{\mathrm{LiAlH}}_4}{\to }$

The property of ${\mathrm{LiAlH}}_4$ is that it​ reduces ester but not double bond.

When an ester reacts with ${\mathrm{LiAlH}}_4$, two alcohols are created, one of which corresponds to the ester's acyl and alkyl portions. 

${\mathrm{H}}_3\mathrm{C}-\mathrm{HC}=\mathrm{CH}-\stackrel{\mathrm{O}}{\mathrm{C}}-{\mathrm{OCH}}_3\stackrel{{\mathrm{LiAlH}}_4}{⟶}{\mathrm{H}}_3\mathrm{C}-\mathrm{HC}=\mathrm{CH}-{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{CH}}_3\mathrm{OH}$

Hence the correct option is (D).