Claim 1: $A,X,B,P$ are concyclic. Construct point $E$ on $AC$ such that $BA,BE$ are symmetric wrt $BC$ We can easily prove that $\Delta DBZ\sim\Delta DCY(A.A)$ $\Rightarrow \frac{DY}{DZ} = \frac{DC}{DB} =\frac{AC}{AB} = \frac{\sin ABC}{\sin ACB}=\frac{\sin EBC}{\sin ECB} = \frac{EC}{EB}$. $(1)$ Moreover, $\angle BEC = 180^\circ - \angle EBC -\angle ACB = \angle ABC - \angle ACB = 180^\circ - 2\angle ADB= \angle ZDY$ $\Rightarrow \Delta BEC\sim\Delta ZDY(S.A.S)$ $\Rightarrow \angle PZD = 180^\circ - \angle DZY = 180^\circ - \angle EBC = \angle ABC$ $\Rightarrow \angle PXD = \angle PZD = \angle ABC\Rightarrow A, X, B, P$ are concyclic. Claim 2: $KP\parallel BZ$: $BZ$ cuts $CY$ at $L$ $\Rightarrow \Delta LBC$ is isosceles $\Rightarrow \frac{PZ}{PY} = \frac{BZ}{CY} =\frac{BX}{CY}=\frac{KB}{KY}\Rightarrow KP\parallel BZ$. Now, back to our problem. The exterior bisector of $\angle BAC$ intersects $BC$ at $F$ $\Rightarrow \frac{KX}{KC} = \frac{BX}{CY} = \frac{BD}{CD}=\frac{FB}{FC}$ $\Rightarrow KF\parallel XB$ $\Rightarrow \angle KFP = \angle XBC =\angle ZBC =\angle KPF$ $\Rightarrow KF = KP$. Otherwise, $\angle FAP = 90^\circ - \angle PAD =90^\circ - \angle XBD=90^\circ - \angle KFP$. Hence, $\angle FAP = \frac{\angle FKP}{2}$, or $A\in (K;KP)$. (Q. E. D)

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I think my solution is similar to the previous post but anyway: My first idea was to get familiar with point $P$ in a natural way. Since $PC$ is bisector of $PX$, $PY$ then we just take a point $Q$ in line $XY$ such that $QP \perp BC$. So we end up with $(Q, U; X, Y) = -1$ where $U$ is the intersection of $XY$ and $BC$. Since we have $BX \| CY$ then we define $V$ as a point in line $BC$ such that $QV \| BX$. Therefore, we have $(V, U; B, C) = -1$ due to the parallelism. So $VA \perp AU$ which implies that $V, Q, A, P$ are concyclic which means that $B, X, A, P$ are concyclic since $VQ \| BX$. Then I realized that it was not necessary to prove that, we just have to use the fact the $PC$ is bisector. Let $W$ be the reflection of $B$ respect to $XY$. Then we have: $\dfrac{KC}{KX} = \dfrac{YC}{BX} = \dfrac{UC}{BU} = \dfrac{AC}{AB} = \dfrac{AC}{AW}$ $\dfrac{KY}{KB} = \dfrac{YC}{BX} = \dfrac{YU}{UX} = \dfrac{YP}{XP} = \dfrac{PY}{PZ}$ So $KA = KP$ is equivalent to $BZ = XW$ which is indeed true since $BZ = BX$ and $BX = XW$ since we just reflected things.