Question about parity operator

Your definition of the parity operator is complete only when it is also specified that $vert alpharangle$ is such that $hat{x}vert alpharangle = alpha vertalpharangle$ where $alphainmathbb{R}$.

Thus, $hat{x}vert alpharangle=alphavert alpharangle$ and $hat{x}vert-alpharangle=-alphavert -alpharangle$. Or, $langlealphaverthat{x}vertalpharangle=alpha$ and $langle-alphaverthat{x}vert-alpharangle=-alpha$. You can easily rewrite the last relation as $langlealphavertmathcal{P}hat{x}mathcal{P}vertalpharangle=-alpha$ and see that thus, $langlealphavertmathcal{P}hat{x}mathcal{P}vertalpharangle=-langlealphaverthat{x}vertalpharangle$.

The point is that the minus sign is brought outside of the ket/bra by the position operator, in particular, notice that $vert -alpha rangle neq -vertalpharangle$. So, to put it in the language that you asked, the minus signs are not canceled out because there is only one position operator and it only brings out one minus sign. The use of the other parity operator is to get a bra which can be sandwiched with the ket to give $1$ (or $delta(0)$ really) rather than $0$ because otherwise you simply get $langlealphaverthat{x}mathcal{P}vert alpharangle=-alphalanglealphavert -alpharangle=0$ (except when $alpha=0$).

It should be added that physically speaking, this is obvious. The expectation value of the position operator should obviously be negative when measured over a state which has been parity reversed. That is literally the meaning of the parity operator.