The answer is $\boxed{\frac{1}{4}}$.

Let $O$ be the center and $P'I$ be a diameter of the circle $\omega$. Then because $PO=UO$ and $IO=P'O$, we find $PIEP'$ is a parallelogram which means $PP'\parallel IE$. It follows that the ares $S_{\triangle{PIE}}=S_{\triangle{P'IE}}$ because they have the same base $IE$ and equal heights.

Now maximizing $S_\triangle{PIE}$ is the same as maximizing $S_{P'IE}$. Because $P'I$ is the diameter of $\omega$, it is clear $S_{\triangle{P'IE}}$ reaches maximum when $IE=P'E$, i.e. $\triangle{P'IE}$ is an isosceles right triangle. This is because the height against the base $P'I$ reaches maximum at this point. Given $PLUM$ is a unit square which means the diameter of $\omega$ equals $1$, we have

$$IE=P'E = \frac{\sqrt{2}}{2}\implies S_{\triangle{P'IE}}=\boxed{\frac{1}{4}}$$

when $P'IE$ is such a triangle.

Finally, we just need to show that a satisfactory isosceles right triangle is obtainable. It is sufficient to show that the length of $IE$ will equals $\frac{\sqrt{2}}{2}$ at some point. To show this, let $X$ be the tangent point where $\omega$ meets $LU$. When $I$ is at the point $X$, the length of $IE=0$. As $I$ moves clock-wise along $\omega$, the length of $IE$ will continuously change. When $IE$ collapse with the diagonal $PU$, its length will equal to the diameter of $\omega$, i.e. $1$. This means that the length of $IE$ will continuously change from $0$ to $1$. Because $\frac{\sqrt{2}}{2}\in (0, 1)$, we conclude it must equal to $\frac{\sqrt{2}}{2}$ at some point.