South African Mathematics Olympiad

$BAP=60^{\circ }$ while $BAC=90^{\circ }$, so $PAC=30^{\circ }$. With $BPA=60^{\circ }$ that means $C=30^{\circ }$, and so $\angle PAC$ is isosceles. Then $BC$ has length $4$, and so by Pythagoras the length of $AC$ is $\sqrt{4^2-2^2}$