IMO Hk TST

We have $BP = \frac{1}{\sqrt{5}}$.
Note that $\angle CBA = 90^\circ$ and $AB = BC$. Since $\angle CBP = \angle BAP$, the line $BC$ is tangent to ($ABP$). As $\angle CBA = 90^\circ$, the centre of this circle must lie on $AB$. Thus, the centre is the midpoint $M$ of $AB$. This implies $\angle APB = 90^\circ$.
Since $\angle BAP = \angle ACP$, the line $AB$ is tangent to ($ACP$). The centre $D$ of this circle must lie on the perpendicular line at $A$ to $AB$, and lie on the perpendicular

bisector of $AC$. Clearly, $D$ is the point such that $ABCD$ is a square. Note that $MD$ is the perpendicular bisector of $AP$. Thus, we have
$$\angle ACP = \frac{1}{2} \angle ADP = \angle ADM.$$
Note that $\tan \angle ADM = \frac{AM}{AD} = \frac{1}{2}$. It follows that
$$BP = AB \sin \angle BAP = \frac{1}{\sqrt{5}}.$$