We know that $$P|\alpha \rangle = |-\alpha \rangle$$

where $$P$$ is the parity operator.

$$\langle \alpha|PxP|\alpha \rangle = - \langle \alpha|x|\alpha \rangle$$

where $$x$$ is the position operator.

How is this correct? Should it not be $$\langle \alpha|x|\alpha \rangle$$ only because the parity operator acts twice?

Your definition of the parity operator is complete only when it is also specified that $$\vert \alpha\rangle$$ is such that $$\hat{x}\vert \alpha\rangle = \alpha \vert\alpha\rangle$$ where $$\alpha\in\mathbb{R}$$.
Thus, $$\hat{x}\vert \alpha\rangle=\alpha\vert \alpha\rangle$$ and $$\hat{x}\vert-\alpha\rangle=-\alpha\vert -\alpha\rangle$$. Or, $$\langle\alpha\vert\hat{x}\vert\alpha\rangle=\alpha$$ and $$\langle-\alpha\vert\hat{x}\vert-\alpha\rangle=-\alpha$$. You can easily rewrite the last relation as $$\langle\alpha\vert\mathcal{P}\hat{x}\mathcal{P}\vert\alpha\rangle=-\alpha$$ and see that thus, $$\langle\alpha\vert\mathcal{P}\hat{x}\mathcal{P}\vert\alpha\rangle=-\langle\alpha\vert\hat{x}\vert\alpha\rangle$$.
The point is that the minus sign is brought outside of the ket/bra by the position operator, in particular, notice that $$\vert -\alpha \rangle \neq -\vert\alpha\rangle$$. 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 $$\langle\alpha\vert\hat{x}\mathcal{P}\vert \alpha\rangle=-\alpha\langle\alpha\vert -\alpha\rangle=0$$ (except when $$\alpha=0$$).