Find $k$ if $P,Q,R,$ and $S$ are points on the sides of quadrilateral $ABCD$ so that \[ \frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = k,  \] and the area of the quadrilateral $PQRS$ is exactly 52% of the area of the quadrilateral $ABCD$.

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1. Given that \( \frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = k \), we need to find \( k \) such that the area of quadrilateral \( PQRS \) is 52% of the area of quadrilateral \( ABCD \).

2. Let the area of quadrilateral \( ABCD \) be denoted as \( [ABCD] \). Then, the area of quadrilateral \( PQRS \) is \( 0.52 \times [ABCD] \).

3. We know that the area of \( PQRS \) can be expressed in terms of the areas of the triangles formed by the points \( P, Q, R, S \) and the vertices of \( ABCD \). Specifically, we can write:
   \[
   [PQRS] = [ABCD] - \left( [SAP] + [PBQ] + [QCR] + [RDS] \right)
   \]

4. Since \( \frac{AP}{PB} = k \), the area of triangle \( SAP \) is proportional to the area of triangle \( SAB \). The ratio of the areas is given by:
   \[
   \frac{[SAP]}{[SAB]} = \frac{k}{k+1}
   \]
   Similarly, for the other triangles:
   \[
   \frac{[PBQ]}{[PBC]} = \frac{k}{k+1}, \quad \frac{[QCR]}{[QCD]} = \frac{k}{k+1}, \quad \frac{[RDS]}{[RDA]} = \frac{k}{k+1}
   \]

5. The areas of triangles \( SAB, PBC, QCD, \) and \( RDA \) are each \(\frac{1}{4}\) of the area of \( ABCD \) because they are formed by dividing the quadrilateral into four parts. Therefore:
   \[
   [SAB] = [PBC] = [QCD] = [RDA] = \frac{1}{4} [ABCD]
   \]

6. Using the area ratios, we can write:
   \[
   [SAP] = \frac{k}{k+1} \cdot \frac{1}{4} [ABCD], \quad [PBQ] = \frac{k}{k+1} \cdot \frac{1}{4} [ABCD], \quad [QCR] = \frac{k}{k+1} \cdot \frac{1}{4} [ABCD], \quad [RDS] = \frac{k}{k+1} \cdot \frac{1}{4} [ABCD]
   \]

7. Summing these areas, we get:
   \[
   [SAP] + [PBQ] + [QCR] + [RDS] = 4 \cdot \frac{k}{k+1} \cdot \frac{1}{4} [ABCD] = \frac{k}{k+1} [ABCD]
   \]

8. Therefore, the area of \( PQRS \) is:
   \[
   [PQRS] = [ABCD] - \frac{k}{k+1} [ABCD] = \left(1 - \frac{k}{k+1}\right) [ABCD] = \frac{(k+1) - k}{k+1} [ABCD] = \frac{1}{k+1} [ABCD]
   \]

9. Given that \( [PQRS] = 0.52 [ABCD] \), we equate and solve for \( k \):
   \[
   \frac{1}{k+1} = 0.52 \implies k+1 = \frac{1}{0.52} = \frac{100}{52} = \frac{25}{13} \implies k = \frac{25}{13} - 1 = \frac{25}{13} - \frac{13}{13} = \frac{12}{13}
   \]

The final answer is \( \boxed{ k = \frac{12}{13} } \)