$\frac{AB}{BC}$ can be $\frac{\sqrt{2}}{2}$ or $\frac{\sqrt{10}}{2}$.
It is well-known that $Q$ is the midpoint of $AC$. We first consider the case when $P$ lies inside $\triangle ABC$. Let $X$ be the projection of $C$ on $BQ$, and let $Y$ be the point on the line $BQ$ such that $XY = XC$, and $X$ lies between $B$ and $Y$.
Note that $\triangle QAP \cong \triangle QCX$, and hence $PQ = XQ$. By the given condition, we have
$$BP = AP + 2PQ = CX + PX = XY + PX = PY.$$
Together with $AP \perp BY$, we know that $\triangle ABY$ is an isosceles triangle, with $AY = AB = AC$. Thus, $AX$ is the perpendicular bisector of $CY$. This implies
$$\angle ACX = \angle AYX = \angle YBA.$$
Therefore, $A$, $B$, $C$, $X$ are concyclic, and $\angle BAC = \angle BXC = 90^\circ$. Conversely, when $\angle BAC = 90^\circ$, one can work backward and show that $BP = AP + 2PQ$. (In that proof, one should define $Y$ as the point on the line $BQ$ such that $AB = AY$, and then show that $XC = XY$.) Thus, this gives one possibility, where
$$\frac{AB}{BC} = \frac{1}{\sqrt{2}}.$$

Next we consider the case when $P$ lies outside $\triangle ABC$. Similarly, let $X$ be the projection of $C$ on $BQ$. As above, we have $\triangle QAP \cong \triangle QCX$, and hence $QX = QP$. By the given condition, we have
$$BP = AP + 2PQ = CX + PX.$$
This implies $BX = CX$. Therefore, $BC = \sqrt{2}CX = \sqrt{2}AP$. Let $PQ = QX = x$ and $CX = AP = y$. Noting that $\angle PBA = \angle XCQ$, we have $\triangle PAB \sim \triangle XQC$.
It follows that $\frac{PA}{XQ} = \frac{PB}{XC}$. This implies
$$\frac{y}{x} = \frac{2x + y}{y}.$$

This can be rewritten as $(2x - y)(x + y) = 0$. Thus, we find that $y = 2x$, and hence
$$\frac{AB}{BC} = \frac{\sqrt{PA^2 + PB^2}}{BC} = \frac{\sqrt{y^2 + (2x+y)^2}}{\sqrt{2}y} = \frac{\sqrt{10}}{2}.$$
When $\frac{AB}{BC} = \frac{\sqrt{10}}{2}$, it is not hard to find that $\frac{BQ}{AB} = \frac{3}{2\sqrt{5}}$, and $\cos \angle CBQ = \frac{1}{\sqrt{2}}$.
Then one can deduce $BX = CX$, and the condition holds.

To conclude, $\frac{AB}{BC}$ can be $\frac{\sqrt{2}}{2}$ or $\frac{\sqrt{10}}{2}$.