b
\(Ph - CH = CH - C = N\xrightarrow{{C{H_3}MgBr/{H_3}{O^ + }}}\) \(\begin{array}{*{20}{c}}
  {Ph - CH - CH - C - NH} \\ 
  {|\,} \\ 
  {\,\,\,\,\,\,C{H_3}\,} 
\end{array}\) \(\xrightarrow{{NaB{H_4},\Delta }}\begin{array}{*{20}{c}}
  {Ph - CH - C{H_2} - C{H_2} - N{H_2}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}\)

In the above reaction, the reactant undergoes \(1,4\) Michael addition first then the formed intermediate undergoes reduction in the presence of \(NaBH _{4}\) to form the desired product.