In an isosceles triangle, one vertex lies on the perpendicular bisector of the opposite side. Thus, either $P$ is the intersection of $A B$ and $\ell$, or $P$ lies on the circle centered at $A$ with radius $A B$, or $P$ lies on the circle centered at $B$ with radius $A B$. Each circle-line intersection has at most two solutions, and the line-line intersection has at most one, giving 5. This can be easily constructed by taking any $\overline{A B}$, and taking $\ell$ that isn't a diameter but intersects both relevant circles twice.