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1 older comment kept for context.
Oluwaseun Adeyemi
For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS).
Thus: \(\bar{PS}^2\) = 52 + 122
\(\bar{PS} = \sqrt{25 + 144}\)
= \(\sqrt{169}\)
\(\bar{PS}\) = 13cm
perimeter = 4 x length of a side
= 4 x 13cm
= 52cm