7. As shown in the figure, in rectangle $A B C D$ with an area of 2, point $E$ is the midpoint of side $A D$. If $\triangle A B E$ and $\triangle D E C$ are folded along $B E$ and $C E$ respectively, such that points $A$ and $D$ coincide at point $P$, then the maximum volume of the tetrahedron $P-E B C$ is . $\qquad$
Given, $E P \perp P B, E P \perp P C \Rightarrow E P \perp$ plane $P B C$. Let $B C=a, A B=b$, then $P B=P C=b, E P=\frac{a}{2} \Rightarrow S_{\triangle P B C}=\frac{1}{2} \cdot a \cdot \sqrt{b^{2}-\frac{a^{2}}{4}}=\frac{1}{4} \cdot a \cdot \sqrt{4 b^{2}-a^{2}}$. Thus, $V_{P-E B C}=V_{E-P B C}=\frac{1}{24} \cdot a^{2} \cdot \sqrt{4 b^{2}-a^{2}}=\frac{1}{24} \sqrt{a^{2}\left(4 a^{2} b^{2}-a^{4}\right)}$ $=\frac{1}{24} \sqrt{a^{2}\left(16-a^{4}\right)}$, Let $t=a^{2}, f(t)=t\left(16-t^{2}\right), f^{\prime}(t)=16-3 t^{2}$, then $f(t)$ is monotonically increasing on $\left(0, \frac{4 \sqrt{3}}{3}\right)$ and monotonically decreasing on $\left(\frac{4 \sqrt{3}}{3}, 4\right)$, i.e., $f(t)_{\max }=\frac{128 \sqrt{3}}{9}$. Therefore, the maximum volume of the tetrahedron $P-E B C$ is $\frac{1}{24} \cdot \frac{8}{3} \cdot \sqrt{2 \sqrt{3}}=\frac{\sqrt[4]{12}}{9}$.