smc

Because $[ABDE]=[△EDC]$, we know that $[△EDC]=\frac{1}{2}[△ABC]$. Because the ratio of their areas is $\frac{1}{2}$, we know that the ratio of their side lengths (and thus their altitudes) is $\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$. Thus, the altitude from $C$ to $\overline{DE}$ has length $1\ast \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$. Thus, because $\overline{DE}\parallel \overline{AB}$, the distance from $P$ to $\overline{AB}$ is $1-\frac{\sqrt{2}}{2}=\boxed{\frac{2-\sqrt{2}}{2}}$.