IMO Hk TST

We have $\angle CAP = 42^\circ$.
Construct a point $D$ such that $\angle DAC = \angle DCA = 72^\circ$, and $B$ lies inside $\triangle DCA$. Since
$$\angle BCD = \angle ACD - \angle ACB = 72^\circ - 60^\circ = 12^\circ = \angle CBP,$$
we have $DC // BP$. Also, note that $BD$ is the perpendicular bisector of $CA$. Therefore, we have
$$\angle CDB = \frac{1}{2} \angle CDA = 18^\circ$$
and
$$\angle PCD = \angle ACD - \angle ACP = 72^\circ - 54^\circ = 18^\circ = \angle CDB.$$
This implies $BDCP$ is an isosceles trapezoid.

Construct a point $E$ such that $DCAE$ is an isosceles trapezoid and $DC//EA$. Since $\triangle EDP$ is the image of reflection of $\triangle ACB$ in the perpendicular bisector of $DC$, it is equilateral. Now, as
$$\angle ADE = \angle CDE - \angle CDA = \angle ACD - \angle CDA = 72^\circ - 36^\circ = 36^\circ$$
and
$$\angle EAD = \angle CDA = 36^\circ = \angle ADE,$$
we have $EA = ED = EP$. It follows that
$$\begin{aligned} \angle PAC &= \angle EAC - \angle EAP \\ &= (180^\circ - \angle ACD) - \left(90^\circ - \frac{1}{2} \angle PEA\right) \\ &= 180^\circ - 72^\circ - 90^\circ + \frac{1}{2}(\angle DEA - \angle DEP) \\ &= 18^\circ + \frac{1}{2}(108^\circ - 60^\circ) \\ &= 42^\circ. \end{aligned}$$