Understanding details of NP-Hard definition
The definition of NP-hardness for a problem $X$ is that every problem in&nbsp;NP reduces to&nbsp;$X$. Informally speaking, "$L$ reduces to&nbsp;$X$" means that if you could solve&nbsp;$X$, you could use that to solve&nbsp;$L$.

So, if $X$&nbsp;is NP-hard, you can reduce all problems in&nbsp;NP, including the ones in&nbsp;P, to&nbsp;$X$. There's no contradiction here: $X$&nbsp;is a hard problem, so saying that you can reduce a problem in&nbsp;P to&nbsp;$X$ is just saying, "If I could solve this hard problem, I could use that to solve a pretty easy problem." Sure, there are easier ways of solving that easy problem (that's why we call it an easy problem!) but all you've done is to show that there's a hard way to solve it, too.


  P is also a part of NP so when we say every problem&nbsp;y which belongs to NP we are saying we can take a problem in&nbsp;P and reduce to some&nbsp;X. Then X&nbsp;is NP-Hard?


Maybe it's just inaccurate writing but you have the implication the wrong way around. If $X$&nbsp;is NP-hard, then you can reduce a problem in&nbsp;P to&nbsp;$X$. The converse (if you can reduce a problem in&nbsp;P to&nbsp;$X$, then $X$&nbsp;is NP-hard) is not true.