Consider, $y=\frac{x-P}{x^2-3x+2}$ and it is provided that
$x^2-3x+2\ne 0$
$x\ne 1, 2$

Now solve the above equation.

$y{x}^2-(3y+1)x+(2y+P)=0$  $\because x\in \mathrm{ℝ},$ 

So $D\ge 0$

$\Rightarrow (3y+1{)}^2-4y(2y+P)\ge 0$

$\Rightarrow y^2+(6-4P)y+1\ge 0$   $\because y\in \mathrm{ℝ},$ so $D\le 0$

$\Rightarrow (6-4P{)}^2-4\le 0$

$\Rightarrow (4P-6+2)(4P-6-2)\le 0$

$\Rightarrow (P-1)(P-2)\le 0$

$P\in [1,2]$ but $P\ne 1, 2$

So, $P\in \left(1,2\right)$