The square $ABCD$ is divided into $8$ equal right triangles and the square $KLMN$, as shown in the figure. Find the area of the square $ABCD$ if $KL = 5, PS = 8$.
[img]https://1.bp.blogspot.com/-B2QIHvPcIx0/X4BhUTMDhSI/AAAAAAAAMj4/4h0_q1P6drskc5zSvtfTZUskarJjRp5LgCLcBGAsYHQ/s0/Yasinsky%2B2020%2Bp1.png[/img]
1. Let \( PK = x \). Since \( PS = 8 \), we have the equation:
   \[
   x^2 + (x + 5)^2 = 64
   \]
   This equation comes from the Pythagorean theorem applied to the right triangle \( PKL \), where \( PK \) and \( KL \) are the legs, and \( PS \) is the hypotenuse.

2. Expand and simplify the equation:
   \[
   x^2 + (x + 5)^2 = 64
   \]
   \[
   x^2 + x^2 + 10x + 25 = 64
   \]
   \[
   2x^2 + 10x + 25 = 64
   \]
   \[
   2x^2 + 10x + 25 - 64 = 0
   \]
   \[
   2x^2 + 10x - 39 = 0
   \]

3. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   a = 2, \quad b = 10, \quad c = -39
   \]
   \[
   x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 2 \cdot (-39)}}{2 \cdot 2}
   \]
   \[
   x = \frac{-10 \pm \sqrt{100 + 312}}{4}
   \]
   \[
   x = \frac{-10 \pm \sqrt{412}}{4}
   \]
   \[
   x = \frac{-10 \pm 2\sqrt{103}}{4}
   \]
   \[
   x = \frac{-10 + 2\sqrt{103}}{4} \quad \text{(since \( x \) must be positive)}
   \]
   \[
   x = \frac{-5 + \sqrt{103}}{2}
   \]

4. The side length of the square \( ABCD \) is \( 5 + 2x \):
   \[
   5 + 2x = 5 + 2 \left( \frac{-5 + \sqrt{103}}{2} \right)
   \]
   \[
   5 + 2x = 5 + (-5 + \sqrt{103})
   \]
   \[
   5 + 2x = \sqrt{103}
   \]

5. The area of the square \( ABCD \) is:
   \[
   (\sqrt{103})^2 = 103
   \]

The final answer is \(\boxed{103}\).