In square $ABCD$ with side length $2$, the inscribed circle $\omega$ intersects $\overline{CD}$ at $M$. Let $E$ be on $\overline{AB}$ with $AE=\dfrac14$, and let $\overline{EM}$ meet $\omega$ again at $P$. What is $AP$?

1. Set $A=(0,0),B=(2,0),C=(2,2),D=(0,2)$, so $\omega$ has center $O=(1,1)$, radius $1$, and meets $\overline{CD}$ at $M=(1,2)$. With $AE=\frac14$, $E=\left(\frac14,0\right)$.
2. Line $EM$ has slope $\dfrac{2-0}{1-1/4}=\dfrac83$, i.e. $y=\dfrac83x-\dfrac23$.
3. Substituting into $(x-1)^2+(y-1)^2=1$ gives $73x^2-98x+25=0$, with roots $x=1$ (point $M$) and $x=\dfrac{25}{73}$ (point $P$).
4. So $P=\left(\dfrac{25}{73},\dfrac{18}{73}\right)$, and $AP=\dfrac{\sqrt{25^2+18^2}}{73}=\dfrac{\sqrt{949}}{73}$.
5. The answer is $\boxed{\dfrac{\sqrt{949}}{73}}\approx0.422$.