@ACuriousMind 's answer is pretty straightforward. As an alternate proof consider the following:
- Note that $\hat P^\mu$ is time-independent$^1$, which means that it commutes with $\hat H$. So it commutes with $\exp[-i\hat HT]\ \forall T\in \mathbb C$.
- We know$^2$ that $|\Omega\rangle \propto \lim_{T\to\infty} \exp[-i\hat HT]|0\rangle$. Using this, its easy to see that$$\hat P^\mu |\Omega\rangle\propto \lim_{T\to\infty} \hat P^\mu\mathrm e^{-i\hat HT}|0\rangle=\lim_{T\to\infty} \mathrm e^{-i\hat HT}\hat P^\mu|0\rangle=0$$where I used $[\hat H,\hat P^\mu]=0$ and $\hat P^\mu|0\rangle=0$.
$^1$ here, $\hat P^\mu$ and $\hat H$ are the total (interacting) momentum-hamiltonian operators. They are time-independent because they are the charges of the Noether current associated to space-time translations.
$^2$ See Quantum Field Theory I, lecture notes by Timo Weigand, page 57.