Inspired by Euler

The expression $OP^2=R(R-2PK)$ looks sort of similar to Euler's formula: $OI^2=R(R-2r)$ . Indeed, another way to interpret the formula is that the power of $I$ with respect to the circumcircle equals $2Rr$ , and this allows a synthetic approach to proving the formula. In fact, I found the following lemma:

Lemma : If $P$ is inside $\triangle ABC$ and satisfies $OP^2=R(R-2PK)$ , where $K$ is its foot of projection on $AB$ , then $QP=QB$ where $Q=AP\cap \odot (ABC)\ne A$ .

Proof : We make use of the interpretation mentioned above, namely $2R\cdot PK=R^2-OP^2=AP\cdot PQ$ Construct $QS$ the diameter of $\odot (ABC)$ . Since $\angle QBS=90^{\circ}=\angle AKP$ and $\angle BSQ=\angle BAP$ , therefore $AKP\sim SBQ\implies$ $AP\cdot BQ=QS\cdot PK=2R\cdot PK=AP\cdot PQ$ Hence $BQ=PQ$ $_{\Box}$

Applying this lemma to the problem, we extend $AD$ to meet $\odot (ABC)$ at $F$ and get $FP=FB$ .Hence $AP=AF-PF=AF-BF$ , which we can definitely calculate:

By Pythagorean theorem: $CD=30, BD=12$ . From $ADC\sim BDF$ , we have $DF=22.5, BF=25.5\implies$ $AP=16+22.5-25.5=\boxed {13}$