In \[\Delta \text{ }ABC,\]
\[AC\text{ }=\text{ }AB\text{ }\left[ Given \right]\]
In this way, \[\angle ABC\text{ }=\angle ACB\][Angles inverse to rise to sides are equal.]
In \[\Delta \text{ }PRC\text{ }and\text{ }\Delta \text{ }PQB,\]
\[\angle ABC\text{ }=\angle ACB\]
\[\angle PRC\text{ }=\angle PQB\text{ }\left[ Both\text{ }are\text{ }correct\text{ }angles. \right]\]
Thus, \[\vartriangle PRC\text{ }\sim\text{ }\vartriangle PQB\] by \[AA\]rule for similitude
Since, relating sides of comparative triangles are corresponding we have
\[PR/PQ\text{ }=\text{ }RC/QB\text{ }=\text{ }PC/PB\]
\[PR/PQ\text{ }=\text{ }PC/PB\]
\[9/15\text{ }=\text{ }12/PB\]
Along these lines,
\[PB\text{ }=\text{ }20\text{ }cm\]