$A$ :



$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} O{\mkern 1mu} } \\
{\,\,\,|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} ||} \\
{C{H_3} - CH - C - OH}
\end{array}$ $+ C{H_3} - N{H_2} \to 'A'\xrightarrow[\Delta ]{}'B'\xrightarrow{\begin{subarray}{l}
{\text{LiAl}}{{\text{H}}_4} \\
{\text{(excess)}}
\end{subarray} }'C'$
The final product $‘C’$ will be
Product $(B)$ is

$(1)$ $C{H_3}C{H_2}N{H_2}$ $(2)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,}\\
{\,C{H_3}C{H_2}NH\,\,\,\,\,\,\,}
\end{array}$
$(3)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{{H_3}C - N - C{H_3}}
\end{array}$ $(4)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{Ph - N - H}
\end{array}$