Assume that $\omega_B$ and $\omega_C$ intersect again at second point $S$ other than $P$. If not, the degenerate case where $\omega_B$ and $\omega_C$ are tangent at $P$ can be dealt similarly. Because $BPSR$ and $CPSQ$ are cyclic, we have $\angle RSP = 180^\circ - \angle PBR$ and $\angle PSQ = 180^\circ - \angle QCP$. Hence, we obtain
$$\angle QSR = 360^\circ - \angle RSP - \angle PSQ = \angle PBR + \angle QCP = \angle CBA + \angle ACB = 180^\circ - \angle BAC,$$
from which it follows that $ARSQ$ is cyclic. This means that $\omega_A, \omega_B$, and $\omega_C$ meet at $S$. Note that this is the statement of Miquel's theorem.
Because $BPSY$ is inscribed in $\omega_B$, we find $\angle XYS = \angle PYS = \angle PBS$. Because $ARXS$ is inscribed in $\omega_A$, we find $\angle SXY = \angle SXA = \angle SRA$. Because $BPSR$ is inscribed in $\omega_B$, we find $\angle SRA = \angle SPB$. Thus, we have $\angle SXY = \angle SRA = \angle SPB$. In triangles $SYX$ and $SBP$, we have $\angle XYS = \angle PBS$ and $\angle SXY = \angle SPB$. Therefore, triangles $SYX$ and $SBP$ are similar to each other which in particular implies
$$\frac{YX}{BP} = \frac{SX}{SP}.$$
In a similar manner, we can show that triangles $SXZ$ and $SPC$ are similar to each other and that
$$\frac{SX}{SP} = \frac{XZ}{PC}.$$
Combining the last two equations yields the desired result.

We consider the configuration shown in the diagram above. We can adjust the proof easily for other configurations. In particular, our proof uses directed angles modulo $180^\circ$.
Let line $RY$ intersect $\omega_A$ again at $T_Y$ (other than $R$). Because $BPYR$ is cyclic, $\angle T_Y YX = \angle T_Y YP = \angle RBP = \angle ABP$. Because $ARXT_Y$ is cyclic, $\angle XT_Y Y = \angle XAR = \angle PAB$. Hence triangles $T_Y YX$ and $ABP$ are similar to each other. In particular, we conclude that
$$\angle YXT_Y = \angle BPA \quad \text{and} \quad \frac{YX}{BP} = \frac{XT_Y}{PA}. \qquad (12)$$
Likewise, if line $QZ$ intersect $\omega_A$ again at $T_Z$ (other than $R$), we can show that triangles $T_ZZX$ and $ACP$ are similar to each other and that
$$\angle T_Z XZ = \angle APC \quad \text{and} \quad \frac{XT_Z}{PA} = \frac{XZ}{PC}. \qquad (13)$$
In the light of the second equalities in (12) and (13), it suffices to show that $T_Z = T_Y$. On the other hand, the first equalities in (12) and (13) imply that $X, T_Y, T_Z$ lie on a line. But this line has only two intersections with $\omega_A$ with $X$ being one of them. Hence we must have $T_Y = T_Z$, completing our proof.

We maintain the configuration and the notations of the second solution. Let $T_1$ denote the intersection of lines $RY$ and $QZ$. Because $BPYR$ and $CPZQ$ are cyclic, we have $\angle T_1YZ = \angle T_1YP = \angle RBP = \angle ABC$ and $\angle YZT_1 = \angle YZQ = \angle PCQ = \angle BCA$. Hence, triangles $T_1YZ$ and $ABC$ are similar to each other. In particular, we have $\angle ZT_1R = \angle ZT_1Y = \angle CAB = \angle QAR$, implying that $AQRT_1$ is cyclic. Therefore, $T_1$ lies on $\omega_A$ and $T_1 = T$. Because $ARXT_1$ is cyclic, $\angle XT_1Y = \angle XT_1R = \angle XAR = \angle PAB$, hence $X$ and $P$ are corresponding points in the similar triangles $T_1YZ$ and $ABC$, from which the desired result follows.