When an alkyne is reduced with $H_2$ in presence of $Pd/BaSO_4$, a cis-alkene is obtained . $H_2/Pt$ will reduce it (alkyne) is an alkane. An alkyne will not be reduced by $NaBH_4$.
A trans-alkene will be formed by reduction of an alkyne by active metal in liquid $NH_3$.
Hence (d) is the correct answer.