Rectangle $ABCD$ has $AB = 8$ and $BC = 13$. Points $P_1$ and $P_2$ lie on $AB$ and $CD$ with $P_1P_2 \parallel BC$. Points $Q_1$ and $Q_2$ lie on $BC$ and $DA$ with $Q_1Q_2 \parallel AB$. Find the area of quadrilateral $P_1Q_1P_2Q_2$.
1. Given a rectangle \(ABCD\) with \(AB = 8\) and \(BC = 13\), we need to find the area of the quadrilateral \(P_1Q_1P_2Q_2\).
2. Points \(P_1\) and \(P_2\) lie on \(AB\) and \(CD\) respectively, with \(P_1P_2 \parallel BC\). Points \(Q_1\) and \(Q_2\) lie on \(BC\) and \(DA\) respectively, with \(Q_1Q_2 \parallel AB\).
3. Let \(X\) be the intersection of \(P_1P_2\) and \(Q_1Q_2\).
4. The area of quadrilateral \(P_1Q_1P_2Q_2\) can be divided into four triangles: \([P_1XQ_1]\), \([Q_1XP_2]\), \([P_2XQ_2]\), and \([Q_2XP_1]\).
5. Each of these triangles can be considered as half of the area of the rectangles formed by extending the sides of the quadrilateral to the sides of the rectangle \(ABCD\).

\[
[P_1Q_1P_2Q_2] = [P_1XQ_1] + [Q_1XP_2] + [P_2XQ_2] + [Q_2XP_1]
\]

6. Each triangle can be expressed as half of the area of the corresponding rectangle:

\[
[P_1XQ_1] = \frac{1}{2}[BP_1XQ_1], \quad [Q_1XP_2] = \frac{1}{2}[CQ_1XP_2], \quad [P_2XQ_2] = \frac{1}{2}[DP_2XQ_2], \quad [Q_2XP_1] = \frac{1}{2}[AQ_2XP_1]
\]

7. Summing these areas, we get:

\[
[P_1Q_1P_2Q_2] = \frac{1}{2}[BP_1XQ_1] + \frac{1}{2}[CQ_1XP_2] + \frac{1}{2}[DP_2XQ_2] + \frac{1}{2}[AQ_2XP_1]
\]

8. Since the sum of the areas of these rectangles is equal to the area of the original rectangle \(ABCD\):

\[
[P_1Q_1P_2Q_2] = \frac{1}{2} \left( [BP_1XQ_1] + [CQ_1XP_2] + [DP_2XQ_2] + [AQ_2XP_1] \right) = \frac{1}{2}[ABCD]
\]

9. The area of rectangle \(ABCD\) is:

\[
[ABCD] = AB \times BC = 8 \times 13 = 104
\]

10. Therefore, the area of quadrilateral \(P_1Q_1P_2Q_2\) is:

\[
[P_1Q_1P_2Q_2] = \frac{1}{2} \times 104 = 52
\]

The final answer is \(\boxed{52}\).