Its a parallelogram
<C+<D=<A+<C
Also
AB||CD
[tex]\\ \rm\Rrightarrow <A+<B=180[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{1}{2}<A+\dfrac{1}{2}<B=\dfrac{1}{2}180[/tex]
[tex]\\ \rm\Rrightarrow <PAB+<ABP=90[/tex]
Using angle sum property
[tex]\\ \rm\Rrightarrow <APB=180-90=90[/tex]
Hence proved
