$\begin{align}

  & \angle NMP=\angle QPA=m\text{   }\!\!\{\!\!\text{ corresponding }\angle \text{s }\!\!\}\!\!\text{ } \\

 & \angle QPA+\angle MPQ=m+p={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ }\angle \text{s on a st}\text{.line }\!\!\}\!\!\text{ } \\

 & \angle PQB=\angle MNQ=n\text{   }\!\!\{\!\!\text{ corresponding }\angle \text{s }\!\!\}\!\!\text{ } \\

 & \angle PQB+\angle NQP=n+q\text{  }\!\!\{\!\!\text{ }\angle \text{s on a st}\text{.line }\!\!\}\!\!\text{ } \\

 & \angle PQB=\angle MPQ\text{    }\!\!\{\!\!\text{ Alternate angles }\!\!\}\!\!\text{ } \\

 & i.e.\text{ }n=p \\

 & \angle NQP=\angle QPA\text{   }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\

 & i.e.\text{ }m=p \\

\end{align}$

From the above only II and IV satisfies the options given. The correct option is D