$A$ and $C$ are fixed points on fixed circles $O_1$ and $O_2$.
Point $B$ is moving on circle $O_1$.
Point $D$ is the intersection of the circle through $A,B,C$ with circle $O_2$.
Point $F,G$ are the intersection of line $BD$ with circles $O_1$ and $O_2$.
Point $P$ is on the line $BD$ such that
$$\tag1\label1
\overrightarrow{PF}⋅\overrightarrow{PG}=\overrightarrow{PB}⋅\overrightarrow{PD}.
$$
Show that the trajectory of $P$ is a hyperbola, which is tangent to the moving line $BD$ and the circles $O_1, O_2$.
This is a partial solution: I showed that the trajectory of $P$ is tangent to the moving line $BD$. But I don't know how to show the trajectory of $P$ is a hyperbola and is tangent to circles $O_1,O_2$.
Make a copy $B',D'$ of $B,D$. Define $P$ be the intersection of $BB'$ and $DD'$. We will show the limit of $P$ is the same point as the point $P$ defined by equation $(1)$.
By this problem we have $$\frac{FF'}{BB'}=\frac{DD'}{GG'}$$ so $$\frac{PF}{PB'}=\frac{PD'}{PG}$$ As $B'\to B,D'\to D$, we get $$\frac{PF}{PB}=\frac{PD}{PG}$$ So this point $P$ is a solution of the equation $(1)$. But there is a unique point on line $BD$ that satisfies the equation $(1)$, since the equation $(1)$ is equivalent to $\frac{PF}{PB}=\frac{FD}{BG}$.
PS: As in my comment, if using unsigned distance, the equation $(1)$ have multiple solutions on line $BD$.