RSA encrypting private keys blinding
I will give the hint:

It is a simple Mathematics rule, given $cp, e$, and $n$ we have this relation:

$(k \cdot p)^e \equiv cp \bmod n$, then we can write this modular relation as;

$k^e \cdot p^e = cp + l\cdot n$, for some $l \in \mathbb{Z}$

$p$ divides $n$ that we already know, and from the equality, $p$ also divides the $np$

Rest left to the op.