We assume that \[ABCD\]be the given cyclic quadrilateral.

\[PA\text{ }=\text{ }PD\text{ }\left[ Given \right]\]

So, \[\angle PAD\text{ }=\angle PDA\text{ }\ldots \ldots \text{ }\left( 1 \right)\]

[Angles opposite to equal sides are equal]

And,

\[\angle BAD\text{ }=\text{ }{{180}^{o}}-\angle PAD\] [Linear pair of angles]

Similarly,

\[\angle CDA\text{ }=\text{ }{{180}^{o}}-\angle PDA\]

\[=\text{ }{{180}^{o}}-\angle PAD\text{ }\left[ From\text{ }\left( 1 \right) \right]\]

As the opposite angles of a cyclic quadrilateral are supplementary,

\[\angle ABC\text{ }=\text{ }{{180}^{o}}-\angle CDA\]

\[=\text{ }{{180}^{o}}-\text{ }({{180}^{o}}-\angle PAD)\text{ }=\angle PAD\]

And, \[\angle DCB\text{ }=\text{ }{{180}^{o}}-\angle BAD\]

\[=\text{ }{{180}^{o}}-\text{ }({{180}^{o}}-\angle PAD)\text{ }=\angle PAD\]

Hence,

\[\angle ABC\text{ }=\angle DCB\text{ }=\angle PAD\text{ }=\angle PDA\]

And it’s is only possible when \[AD\text{ }||\text{ }BC.\]