Solution:

Since $\text{XAD= XBC}$ and $\text{XDP= XCP}$, it holds that $\triangle XAD \sim \triangle XBC$.

Let the bisector $s_{X}$ of angle $AXC$ intersect the circles $PAB$ and $PCD$ at points $K$ and $R$, and let the bisector $s_{P}$ of angle $APC$ intersect the circles $PAB$ and $PCD$ at points $L$ and $S$. The line $s_{P}$ passes through the center $I$ of circle $k$ and it holds that $LA=LB=LI$ and $SC=SD=SI$.

Since $\text{ILK= PXK= PXR= ISR}$, it holds that $KL \| RS$. Furthermore, we have $\text{RXS= RXC- SPC= 1 2 ( AXC- APC)= 1 2 BXC}$ and, similarly, $\text{LXK= 1 2 BXC}$. It follows that equal inscribed angles correspond to the chords $KL$ and $RS$ in circles $PAB$ and $PCD$, as well as to the chords $LB$ and $SD$, so $\frac{KL}{RS}=\frac{LB}{SD}=\frac{LI}{SI}$. It follows that $\triangle IKL \sim \triangle IRS$, so point $I$ lies on line $KR$, which is the bisector of angle $AXC$. The claim of the problem follows immediately.

Second solution. Let us denote by $U$ and $V$, respectively, the intersections of the bisectors of angles $AXD$ and $BXC$ with $AD$ and $BC$. As in the first solution, $\triangle XAD \sim \triangle XBC$, from which $\text{XUP= XVP}$, so points $X, P, U$ and $V$ lie on the same circle $\gamma$. We

will prove that point $I$ is also on this circle and that $IU=IV$. It will follow that $I$ belongs to the bisector of angle $UXV$, which is at the same time the bisector of angle $AXC$.

Let $M$ and $N$ be, respectively, the points of tangency of circle $k$ with sides $AD$ and $BC$. Let us denote $AM=a, BN=b, CN=c$ and $DM=d$. Then $AB=a+b, CD=c+d$ and $AU:UD=(a+b):(c+d)$, from which we find $AU=\frac{a+b}{a+b+c+d} \cdot AD=\frac{(a+b)(a+d)}{a+b+c+d}$; similarly we have $BV=\frac{(b+a)(b+c)}{a+b+c+d}$. It follows that $AM-AU=BV-BN=\frac{ac-bd}{a+b+c+d}$, so $MU=NV$ and triangles $IMU$ and $INV$ are congruent and identically oriented. Therefore, $IU=IV$ and $\text{UIV= MIN=180 - VPU}$, i.e., $I$ is the midpoint of arc $UV$ of circle $PXUV$.

Third solution. The following statement from projective geometry is known.
- Desargues's involution theorem. Conic $\gamma$ is circumscribed about quadrilateral $\overline{A B C D}$. Line $\ell$ intersects $AB, CD, BC, DA, AC, BD$ respectively at points $X_{1}$, $X_{2}, Y_{1}, Y_{2}, Z_{1}, Z_{2}$, and intersects the conic $\gamma$ at $W_{1}$ and $W_{2}$. Then there exists an involution on line $\ell$ that maps $X_{1} \leftrightarrow X_{2}, Y_{1} \leftrightarrow Y_{2}, Z_{1} \leftrightarrow Z_{2}$ and $W_{1} \leftrightarrow W_{2}$.

The dual statement (obtained by polar mapping with respect to $\gamma$) is as follows:
- Conic $\gamma$ is inscribed in quadrilateral $ABCD$ in which $AD \cap BC=\{P\}$ and $AB \cap CD=\{Q\}$. Lines $XU$ and $XV$ are tangents from an arbitrary point $X$ to $\gamma$. Then there exists an involution on the pencil of lines through $X$ that maps $XA \leftrightarrow XC, XB \leftrightarrow XD, XP \leftrightarrow XQ$ and $XU \leftrightarrow XV$.

In our case, angles $AXC, BXD$ and $PXQ$ have a common bisector $s$, so the mentioned involution is precisely the axial symmetry with respect to $s$. It follows that the two tangents from $X$ to the circle (conic) $k$ are symmetric with respect to $s$.