Let $O$ be the center of $A B C$. First, we draw in the circumcircle of $A B C$ and the circumcircle of $B O C$, labeled $\omega_{1}$ and $\omega_{2}$, respectively. Note that $\omega_{1}$ is the reflection of $\omega_{2}$ over $B C$ and that $P$ lies on $\omega_{2}$. Now, let $P_{C}$ be the second intersection of ray $C P$ with $\omega_{1}$. Additionally, label the second intersections of ray $A P$ with $\omega_{1}$ and $\omega_{2}$ be $M$ and $X$, respectively. Lastly, let $A^{\prime}$ be the diametrically opposite point from $A$ on $\omega_{1}$. We first note that $A^{\prime}$ is the center of $\omega_{2}$. Thus, $A^{\prime}$ lies on the perpendicular bisector of segment $P X$. But since $A A^{\prime}$ is a diameter of $\omega_{1}$, this also means that the midpoint of $P X$ lies on $\omega_{1}$. This implies that $M$ is the midpoint of $P X$. From a simple angle chase, we have $\angle P_{C} P B=180-\angle B P C=60^{\circ}$. Also, $\angle B P_{C} C=\angle B A C=60^{\circ}$. Therefore, we find that triangle $B P P_{C}$ is equilateral with side length $B P$. Now we begin computations. By Law of Cosines in triangle $B P C$, we see that $B P^{2}+C P^{2}+B P \cdot C P=$ $B C^{2}=A B^{2}$. However, we can rewrite this as $$A B^{2}=B P^{2}+C P^{2}+B P \cdot C P=(B P+C P)^{2}-B P \cdot C P=2 \cdot A P^{2}-B P \cdot C P$$ To find an equation for $\frac{A P}{A B}$, it suffices to simplify the expression $B P \cdot C P$. Since $B P P_{C}$ is equilateral, we can proceed through Power of a Point. By looking at $\omega_{1}$, we see that $$B P \cdot C P=P P_{C} \cdot C P=A P \cdot P M=\frac{1}{2} \cdot A P \cdot A X$$ Then, from Power of a Point on $\omega_{2}$, we see that $$\frac{1}{2} \cdot A P \cdot A X=\frac{1}{2} \cdot A P \cdot(A X-A P)=\frac{1}{2} \cdot A P \cdot A X-\frac{1}{2} \cdot A P^{2}=\frac{1}{2}\left(A B^{2}-A P^{2}\right)$$ Combining everything, we find that $B P \cdot C P=\frac{1}{2}\left(A B^{2}-A P^{2}\right)$ which means that $$A B^{2}=2 \cdot A P^{2}-\frac{1}{2}\left(A B^{2}-A P^{2}\right) \Longrightarrow \frac{5}{2} A B^{2}=\frac{3}{2} A P^{2} \Longrightarrow \frac{A P}{A B}=\frac{\sqrt{15}}{5}$$
Point $P$ lies inside equilateral triangle $A B C$ so that $\angle B P C=120^{\circ}$ and $A P \sqrt{2}=B P+C P$. $\frac{A P}{A B}$ can be written as $\frac{a \sqrt{b}}{c}$, where $a, b, c$ are integers, $c$ is positive, $b$ is square-free, and $\operatorname{gcd}(a, c)=1$. Find $100 a+10 b+c$.