PA=PB PA*PG=PH*PB => PG=PH => AG=BH FG/OP=AG/AP=BH/BP => FG=EH => ⊙E ⊙F are congruence ⊙F is the reflection of ⊙E W.R.T OP => AB ⊥ OP

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Let the line $POC$ meet $(O)$ at $S$. From an inversion with centre $S$ and appropriate radius, we get two tangential circles $(K_1), (K_2)$(The centres need not map to the centre of the image circle) both homothetic to the original circles and their tangency point is $C'$. $(O)$ maps to a common tangent $l$ of $(K_1)$ and $(K_2)$ and $l$ intersects $(K_1)$ and $(K_2)$ at $A'$ and $B'$ respectively. $P'$ lies in the image of the line $SCP$ which means $SC'$ intersects the line $l$ at $P'$ and also, point $O'$ lies on $SC'P'$. It suffices to prove that $SP'$ is perpendicular to $A'B'$. Consider quadrilateral $K_1C'P'A'$ where $K_1C'=K_1A'$ and $\angle K_1A'P'=\angle K_1C'P'=90^{\circ}\Longrightarrow K_1C'P'A'$ is a square$\Longrightarrow SP'\parallel C'P'\parallel K_1A'\perp A'B'$ as required. [Edit: Wrong as pointed out]

Maybe I misunderstood something, but PA = PB and OA = OB mean that PO is the perpendicular bisector of AB, which would finish the problem.

gouthamphilomath, I actually don't understand your solution. It seems to me that you didn't use $PA = PB$ at all. Also, I don't understand your proof that that quadrilateral is a square. Finally, notice that $AB \perp OP$ in general is not equivalent to $A'B' \perp O'P'$, because lines may invert into circles (that is, $A'B'$ is not the inverse of $AB$) Anyway, the problem as it is sounds weird because all the tangent circles turn out to be useless to the solution.

$\triangle PAB$ is isosceles, its circumcenter lies on the angle bisector of $\angle APB$, done. Best regards, sunken rock