Is P contained in NP-hard?
NP-hard has all problems that can be reduced in polynomial time from an NP to them.


Not quite: NP-hard consists of all problems to which every NP problem reduces.

Suppose $X\in \mathsf{P}$. If $\mathsf{P}\not=\mathsf{NP}$, then $SAT$ (for example) is not reducible to $X$. So $X$ is not $\mathsf{NP}$-hard: there are some problems in $\mathsf{NP}$ which do not reduce to $X$.

Note that the class of problems to which some problem in $\mathsf{NP}$ reduces is ... the class of all problems whatsoever! This is because (as you observe) once we discard the "edge cases" $\emptyset$ and $\mathbb{N}$, every problem in $\mathsf{P}$ is also in $\mathsf{NP}$, and problems in $\mathsf{P}$ are trivially reducible$^1$ to everything.



$^1$With respect to polynomial-time many-one reducibility. Change the reducibility, and you potentially change the situation. E.g. $\mathsf{P}$ is $\mathsf{NP}$-hard with respect to exponential-time many-one reducibility, since with respect to this reducibility all problems in $\mathsf{NP}$ are trivial.


The "many-one" bit is why $\emptyset$ and $\mathbb{N}$ are problematic: no nonempty set is many-one reducible to $\emptyset$, and dually no no set other than $\mathbb{N}$ is many-one reducible to $\mathbb{N}$. This annoyance goes away with more intricate reducibilities, like (variations of) Turing reducibility: literally every problem in $\mathsf{P}$ is reducible to everything with respect to polynomial-time Turing reducibility.


In complexity theory, the default reducibility notion is polynomial-time many-one reducibility. So if we just say "$\mathsf{NP}$-hard" we mean with respect to that reducibility - in which case "there is an $\mathsf{NP}$-hard problem in $\mathsf{P}$" is equivalent to $\mathsf{P=NP}$.