Is this even possible to have Integer value?

L.H.S. = $\frac{1}{x}-[\frac{1}{x}]=\frac{1}{x}$ , since $x$ lies between $\sqrt{2},\sqrt{3}$ , and its reciprocal is less than $1$ .

R.H.S. = $x^2-[x^2]$ . Now, $2<x^2<3\Rightarrow[x^2]=2$

Therefore, simplifying we get $\frac{2x+1}{x}=x^2$ . Adding and multiplying both sides by $x$ , we get $(x+1)^2=x^2(x+1)$ . Since $x+1\neq0\Rightarrow x^2-x-1=0$

Using the quadratic equation, we get $x=\frac{1+\sqrt{5}}{2}$ ,as the other root is discarded for being outside of the domain.

Thus, the value of $P$ with $x=\frac{1+\sqrt{5}}{2}$ , is $\boxed{5}$

Sorry if my solution is too brief.