5. The area of convex quadrilateral $A B C D$ is 1, and its diagonals intersect at $P$. The centroids of $\triangle P A B, \triangle P B C, \triangle P C D, \triangle P D A$ are $M_{1}, M_{2}, M_{3}, M_{4}$, respectively. Then the area of quadrilateral $M_{1} M_{2} M_{3} M_{4}$ is
5. Let the midpoints of $P B$ and $P D$ be $K$ and $H$ respectively. Connect $A K$, $A U$, $C K$, $C H$, then $M_{1}, M_{2}, M_{3}, M_{4}$ are the trisection points of $A K$, $C K$, $C H$, $A H$ respectively.
$$
\begin{aligned}
& \because S_{A B C D}=1, \therefore S_{A K C H}=\frac{1}{2} . \\
\because & \frac{A M_{1}}{A K}=\frac{A M_{4}}{A H}=\frac{2}{3}, \\
\therefore & S_{A N_{1} N_{4}}=\frac{4}{9} S_{A K H} .
\end{aligned}
$$

Similarly, $S_{\mathrm{CM}_{2} \mathrm{~N}_{3}}=\frac{4}{9} S_{C K H}$.
$$
\begin{aligned}
\therefore & S_{A N_{1} M_{4}}+S_{C N_{2} M_{3}} \\
& =\frac{4}{9} S_{A K C H} .
\end{aligned}
$$

Similarly, $S_{K M_{1} M_{2}}+S_{H M_{3} M_{4}}=\frac{1}{9} S_{A K C H}$.
Thus, $S_{N_{1} N_{2} M_{3} M_{4}}=\frac{4}{9} S_{A K C H}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}$.