You can prove by induction on p. After that, you can prove by induction on m and a nested induction on n.

The idea is in the in the nested induction step you assume s(m) * s(p) = s(n) * s(p), i.e., m * s(p) + s(p) = n * s(p) + s(p), so s(p) can be canceled and the claim follows by the induction hypothesis for m.