Solution:

The distances from an interior point to the parallel lines $DA$ and $BC$ add up to the distance between them, the height of the rhombus. So the height is $9 + 16 = 25,$ and the incircle, tangent to both lines, has radius $\frac{25}{2}.$ Center the incircle at the origin with $DA:\ y = \frac{25}{2}$ and $BC:\ y = -\frac{25}{2}.$ Then $P$ has $y$-coordinate $\frac{25}{2} - 9 = \frac{7}{2},$ and $x^2 + \left(\frac{7}{2}\right)^2 = \left(\frac{25}{2}\right)^2$ gives $x = \pm 12.$

Let $\alpha = \angle BAD.$ Line $AB$ is tangent to the incircle and makes angle $\alpha$ with the horizontal, so (orienting the figure suitably) it is $x \sin\alpha + y \cos\alpha = -\frac{25}{2},$ and interior points satisfy $x\sin\alpha + y\cos\alpha + \frac{25}{2} \gt 0.$ The condition $\operatorname{dist}(P, AB) = 5$ reads $x \sin\alpha + \frac{7}{2}\cos\alpha + \frac{25}{2} = 5.$ For $x = 12$ the left side exceeds $\frac{25}{2},$ so $x = -12,$ and the equation becomes $24\sin\alpha - 7\cos\alpha = 15.$

Substituting $7\cos\alpha = 24\sin\alpha - 15$ into $\sin^2\alpha + \cos^2\alpha = 1$ yields $625\sin^2\alpha - 720\sin\alpha + 176 = 0,$ so $\sin\alpha = \frac{4}{5}$ or $\frac{44}{125}.$ The root $\frac{44}{125}$ makes $\cos\alpha = \frac{24\sin\alpha - 15}{7}$ negative, contradicting $\angle BAD \lt 90^\circ.$ So $\sin\alpha = \frac{4}{5},$ the side length is $\frac{25}{\sin\alpha} = \frac{125}{4},$ and the perimeter is $4 \cdot \frac{125}{4} = 125.$