jmc

Note that $A$ is the focus of the parabola $y^2=4x,$ and the directrix is $x=-1.$ Then by definition of the parabola, the distance from $P$ to $A$ is equal to the distance from $P$ to the line $x=-1.$ Let $Q$ be the point on $x=-1$ closest to $P,$ and let $R$ be the point on $x=-1$ closest to $B.$



Then by the triangle inequality, $$AP+BP=QP+BP\ge BQ.$$By the Pythagorean Theorem, $BQ=\sqrt{B{R}^2+Q{R}^2}\ge BR=6.$

Equality occurs when $P$ coincides with the intersection of line segment $\overline{BR}$ with the parabola, so the minimum value of $AP+BP$ is $\boxed{6}.$