7. In a convex quadrilateral $A B C D$, $\overrightarrow{B C}=2 \overrightarrow{A D}$. Point $P$ is a point on the plane of the quadrilateral, satisfying $\overrightarrow{P A}+2020 \overrightarrow{P B}+\overrightarrow{P C}+2020 \overrightarrow{P D}=\mathbf{0}$. Let $s, t$ be the areas of quadrilateral $A B C D$ and $\triangle P A B$, respectively. Then $\frac{t}{s}=$ $\qquad$ .
As shown in the figure, let the midpoints of $A B, B D, A C, C D$ be
$$
E, N, M, F \text {, and }|A D|=2,|B C|=4 \text {, }
$$

then $|E N|=|N M|=|M F|=1$,
$$
\begin{array}{l}
\overrightarrow{P A}+\overrightarrow{P C}+2020 \overrightarrow{P B}+2020 \overrightarrow{P D}=\mathbf{0} \\
\Rightarrow \overrightarrow{P M}+2020 \overrightarrow{P N}=\mathbf{0} \Rightarrow \frac{|P N|}{|P M|}=\frac{1}{2020} \\
\Rightarrow|P N|=\frac{1}{2021} \Rightarrow \frac{S_{\triangle P A B}}{S_{\triangle F A B}}=\frac{1+\frac{1}{2021}}{3}=\frac{674}{2021} .
\end{array}
$$

Also, $S_{\triangle F A B}=\frac{1}{2}\left(S_{\triangle C A B}+S_{\triangle D A B}\right)=\frac{1}{2}\left(S_{\triangle A B C}+S_{\triangle D A C}\right)=\frac{1}{2} S_{A B C D}$,
so $\frac{t}{s}=\frac{S_{\triangle P A B}}{S_{A B C D}}=\frac{S_{\triangle P A B}}{2 S_{\triangle F A B}}=\frac{1}{2} \times \frac{674}{2021}=\frac{337}{2021}$.