Correct option is (C) right

In triangles \(\triangle PAQ\;and\;\triangle PBQ,\)

PA = PB,       (Given)

AQ = BQ,      (Given)

PQ = PQ       (Common side)

\(\therefore\) \(\triangle PAQ\cong\triangle PBQ\)    (By SSS congruence criteria))

\(\therefore\) \(\angle APQ=\angle BPQ\)    (By corresponding property of congruence triangles)

\(\Rightarrow\) \(\angle APO=\angle BPO\)   _______(1)

Now, in triangles \(\triangle APO\;and\;\triangle BPO,\)

AP = BP    (Given)

\(\angle APO=\angle BPO\)       (From (1))

OP = OP      (Common side)

\(\therefore\) \(\triangle APO\cong\triangle BPO\)   (By SAS congruence criteria)

\(\therefore\) \(\angle AOP=\angle BOP\) _______(2)  (By corresponding property of congruence triangles)

Since, \(\angle AOP\;and\;\angle BOP\) forms a linear pair.

\(\therefore\) \(\angle AOP+\angle BOP\) = \(180^\circ\)

\(\Rightarrow\) \(2\angle AOP\) = \(180^\circ\)    \((\because\) \(\angle AOP=\angle BOP\)   (From (2)))

\(\Rightarrow\) \(\angle AOP\) = \(\frac{180^\circ}2=90^\circ\)

\(\Rightarrow\) \(\angle POA\) = \(90^\circ\)

Hence, \(\angle POA\) is a right angle.

Correct option is  C) right