smc

Call the midpoint of $\overline{AB}$ point $N.$ Draw in $\overline{NM}$ and $\overline{NP}.$ Note that $\angle NPM=90^{\circ }$ due to Thales's Theorem. Using the Pythagorean theorem, $AM=\frac{\sqrt{5}}{2}.$ Now we just need to find $AP$ using similar triangles $△APN\sim △ANM:$ $$\begin{array}{rl}\frac{AP}{AN}& =\frac{AN}{AM}\\ \frac{AP}{1/2}& =\frac{1/2}{\sqrt{5}/2}\\ AP& =\boxed{\frac{\sqrt{5}}{10}}.\end{array}$$