We consider the configuration shown on the left below. Let $O_B$ and $\omega_B$ ($O_C$ and $\omega_C$) denote the circumcenter and circumcircle of triangle $ABP$ ($ACP$) respectively. Line $ST$, with $S$ on $\omega_B$ and $T$ on $\omega_C$, is one of the common tangent lines of the two circumcircles. Point $X$ lies on segment $ST$. Point $Y$ lies on the other common tangent line.

We will start with the following simple and well known geometry facts.
Let $M$ be the intersection of segments $XY$ and $O_B O_C$. By symmetry, $M$ is the midpoint of both segments $AP$ and $XY$, and line $O_B O_C$ is the perpendicular bisector of segments $XY$ and $AP$. By the power of a point theorem,
$$XS^2 = XA \cdot XP = XT^2 \quad \text{and} \quad X \text{ is the midpoint of segment } ST. \qquad (14)$$
We claim that triangles $ABC$ and $AO_B O_C$ are similar to each other, which is the Salmon theorem. Indeed, $\angle ABC = \angle MO_B A = \angle O_C O_B A$, because each angle is equal to half of the angular size of arc $\widehat{AP}$ of $\omega_B$. Likewise, $\angle O_B O_C A = \angle C$. In particular, we have
$$\frac{AB}{AO_B} = \frac{BC}{O_B O_C} = \frac{CA}{O_C A}. \qquad (15)$$

Set $AB = c$, $BC = a$, and $CA = b$. We claim that it suffices to show the following key fact:
$$1 - \left(\frac{PA}{XY}\right)^2 = \frac{BC^2}{(AB + AC)^2} = \frac{a^2}{(b+c)^2}. \qquad (16)$$
Assuming (16), the given condition in the problem becomes
$$\frac{PB \cdot PC}{AB \cdot AC} = \frac{a^2}{(b+c)^2} \quad \text{or} \quad PB \cdot PC = \frac{a^2bc}{(b+c)^2}. \qquad (17)$$
We claim there are precisely two points $P_1$ and $P_2$ (on segment $BC$) satisfying (17). Write (17) as
$$\frac{a^2bc}{(b+c)^2} = PB \cdot PC = PB \cdot (a - PB),$$
which is a quadratic equation in $PB$, so there are at most two solutions. We now exhibit these solutions explicitly. Construct $P_1$ so that $AP_1$ is the bisector of $\angle BAC$, and let $P_2$ be the reflection of $P_1$ across the midpoint of segment $BC$. Indeed, by the angle-bisector theorem, $P_2C = P_1B = \frac{ac}{b+c}$ and $P_2B = P_1C = \frac{ab}{b+c}$, from which (17) follows for $P_1$ and $P_2$.
It remains now to establish (16); we present two different approaches.

First approach: (By Titu Andreescu and Cosmin Pohoata) Rays $O_B X$ and $O_C T$ meet in $W$. Because of (14) and $O_B S \parallel O_C T$, triangles $O_B S X$ and $W T X$ are congruent to each other. Hence $O_B X = X W$ and triangles $O_B X O_C$ and $W X O_C$ have the same area. Note that $X M$ and $X T$ are altitudes in triangles $O_B X O_C$ and $W X O_C$, respectively. Hence, we obtain
$$\frac{XY \cdot O_B O_C}{4} = \frac{XM \cdot O_B O_C}{2} = \frac{XT \cdot O_C W}{2} = \frac{ST \cdot (O_C T + TW)}{4} = \frac{ST \cdot (O_C T + O_B S)}{4}.$$
By (15), we can write the above equation as
$$\frac{XY}{ST} = \frac{O_C T + O_B S}{O_B O_C} = \frac{O_C A + O_B A}{O_B O_C} = \frac{AB + AC}{BC} \quad \text{or} \quad \frac{XY^2}{ST^2} = \frac{(b+c)^2}{a^2}. \qquad (18)$$

Note that $O_B S T O_C$ is a right trapezoid. Let $U$ be the foot of the perpendicular from $O_C$ on $O_B S$. We have
$$ST^2 = UO_C^2 = O_B O_C^2 - O_S U^2 = O_B O_C^2 - (O_B S - O_C T)^2 = O_B O_C^2 - (O_B A - O_C A)^2.$$

By (15), we can write the above equation as
$$ST^2 = \frac{O_B O_C^2}{BC^2} (BC^2 - (BA - CA)^2) = \frac{O_B O_C^2}{BC^2} (a^2 - (b-c)^2) = \frac{O_B O_C^2}{BC^2} (a+b-c)(a-b+c). \quad (19)$$

Multiplying (18) and (19) together gives
$$XY^2 = \frac{O_B O_C^2}{BC^2} \cdot \frac{(a+b-c)(a-b+c)(b+c)^2}{a^2}. \quad (20)$$
By (15), this implies
$$\frac{ST}{XY} = \frac{O_B O_C}{O_B S + O_C T} = \frac{O_B O_C}{O_B A + O_C A} = \frac{BC}{BA + CA} = \frac{a}{b+c}. \qquad (23)$$

Let $h_a$ denote the length of the altitude from $A$ to side $BC$ in triangle $ABC$. Then $h_a$ and $AM$ are corresponding parts in similar triangles $ABC$ and $AO_B O_C$, and so
$$\frac{AM}{h_a} = \frac{AO_B}{AB} = \frac{O_B O_C}{BC}.$$
It is clear that (16) follows from (22) and (23).