Solution:

Answer: 2 The idea is similar to the proof that the angle bisectors concur or that the perpendicular bisectors concur. Assume WLOG that $BC > AB > CA$. Note that $h_{a}$ and $h_{b}$ are both hyperbolas. Therefore, $h_{a}$ and $h_{b}$ intersect in four points (each branch of $h_{a}$ intersects exactly once with each branch of $h_{b}$). Note that the branches of $h_{a}$ correspond to the cases when $PB > PC$ and when $PB < PC$. Similarly, the branches of $h_{b}$ correspond to the cases when $PC > PA$ and $PC < PA$.
If either $PA < PB < PC$ or $PC < PB < PA$ (which each happens for exactly one point of intersection of $h_{a}$ and $h_{b}$), then $|PC - PA| = |PC - PB| + |PB - PA| = |AB - AC| + |BC - BA| = |BC - AC|$, and so $P$ also lies on $h_{c}$. So, exactly two of the four points of intersection of $h_{a}$ and $h_{b}$ lie on $h_{c}$, meaning that $h_{a}$, $h_{b}$, and $h_{c}$ concur in two points.