Does a problem stay NP-hard when an additional constraint is added to it?
For general $f$ you can't conclude that $P$ is NP-hard just from the fact that $P'$ is NP-hard. In order to do so, you would need to come up with a reduction from $P'$ to $P$. For such a reduction to work, you would need that for every value of $A$ there be a value of $B$ such that the two systems of constraints are equivalent. Unfortunately this is not the case.