vacuum - Does $langleOmega|mathrm{e}^{mathrm{i}Px}=langleOmega|mathrm{e}^{mathrm{i}0x}$? $(langleOmega| =$ ground state of the interacting theory)

Let $\langle\Omega|$ be the ground state of an interacting theory, just as Peskin & Schroeder(PS) describes on page 82 and page 213. On page 213 PS do the following

$$\tag{1}\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle = \langle\Omega|\exp(+\mathrm{i}Px)\phi(0)\exp(-\mathrm{i}Px)|\lambda_{\vec{p}}\rangle \\ =\langle\Omega|\phi(0)|\lambda_{0}\rangle\exp(-\mathrm{i}px)\bigg|_{p^0=E_{\vec{p}}} $$

In the above $|\lambda_0\rangle$ is an eigenstate of the full interacting hamiltonian $H$. The ket $|\lambda_{\vec{p}}\rangle$ is the boost of $|\lambda_0\rangle$ with momentum $\vec{p}.$ Furthermore $E_{\vec{p}} = \sqrt{p^2+m_\lambda^2}$ and $\vec{P}|\lambda_0\rangle = 0.$

I assume PS used $$\tag{2}\langle\Omega|\mathrm{e}^{\mathrm{i}Px}=\langle\Omega|\mathrm{e}^{\mathrm{i}0x}$$ in $(1)$.

Is $(2)$ true?

I thought $\langle0|\mathrm{e}^{\mathrm{i}Px}=\langle0|\mathrm{e}^{\mathrm{i}0x}$ where $\langle0|$ is the ground state of the free theory.

I hope I have provided enough info for the question to make sense.