quantum mechanics - Action of Parity operator on Impulse representation

Is my derivation of the action of the parity operator $\mathbb{P}$ on the $|p\rangle$ representation correct?

$$\left( \mathbb{P}\tilde\psi \right)(p)= - \tilde\psi (p).$$

Obtained from

$$\left( \mathbb{P}\tilde\psi \right)(p) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}px}(\mathbb{P}\psi)(x)= $$

$$ = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}px}\psi(-x)=$$

$$= -\frac{1}{\sqrt{2\pi\hbar}}\int_{\infty}^{-\infty} dx \, e^{-\frac{i}{\hbar}(-p)x}\psi(x)=-\tilde\psi(p).$$

$$= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}(-p)x}\psi(x)=\tilde\psi(-p).$$