There is equivalence in an NP-hardness proof or not?
You only need to show that solving&nbsp;$P_1$ allows you to solve&nbsp;$P_2$ (and, hence, every other problem in NP).

$P_1$ might not be in NP so it's not necessarily the case that solving&nbsp;$P_2$ allows you to solve&nbsp;$P_1$.  On the other hand, if $P_1$ is in NP, then you already know that solving&nbsp;$P_2$ allows you to solve&nbsp;$P_1$, by definition of NP-completeness.