The short answer: Peskin & Schroeder assume that the vacuum is invariant under translations and Lorentz transformations (which implies $P^\mu|\Omega\rangle=0$). Alternatively, translation invariance implies $\vec P|\Omega\rangle=0$ in all reference frames, and combining this with invariance under boosts gives $P^0|\Omega\rangle=0$.

The less short answer: Since P&S is perturbative up to chapter 7, one thing you could ask is this: what interactions, when added perturbatively to a given free theory, will automatically lead to the interacting ground state $|\Omega\rangle$ being Lorentz invariant. Perturbation theory assumes that the interacting ground state $|\Omega\rangle$ overlaps with the free vacuum $|0\rangle$, i.e. $\langle\Omega|0\rangle\neq 0$, so let's set $|\Omega\rangle=a|0\rangle+b|\Delta\rangle$ where $\langle \Delta|0\rangle=0$ and $a\neq 0$. By definition of $|0\rangle$ and the 4-momentum for the free theory $P_0^\mu$, we have $P_0^\mu|0\rangle=0$. The 3-momentum operator for the interacting theory is usually the same as that for the free theory (certainly in the absence of derivative couplings), so let's first suppose $\vec P|0\rangle=0$. In order to have $\vec P|\Omega\rangle = \vec p|\Omega\rangle$ with $\vec p\neq 0$ we would need $\vec P|\Delta\rangle \propto |0\rangle$. This is impossible, however, because the orthogonal complement of $|0\rangle$ is preserved by $P_0^\mu$ ($\vec P$ doesn't change particle number). Together with Lorentz invariance, this implies that $P^\mu|\Omega\rangle=0$ when $\vec P=\vec P_0$ perturbatively in the coupling.

One example where $\vec P \neq \vec P_0$ is in an effective field theory for the dynamics of electrons that ignores photons. Here, there is an ambiguity in the ground state from the choice of background photons that could originate from a distant source. If the background field of photons is sufficiently strong, then the 'lowest energy state'* could be a thermalized 'gas' of electron-positron pairs. The 'free' theory for electrons (Dirac fermions) is Lorentz-invariant, but the interacting theory needs to include terms that account for things like finite propagation times (if the gas is sufficiently dilute). Essentially, there is momentum stored in photons that isn't accounted for in the free Fock space of electrons.

In general, the ground state can differ from the (interacting) vacuum state, spontaneously breaking Lorentz symmetry and leading to a preferred frame. Spontaneous symmetry breaking of this type is necessary if $P^\mu|\Omega\rangle\neq 0$, and the preferred frame is the one in which 3-momentum vanishes. Lorentz invariance can be broken in other ways too, e.g. through a vacuum with nonzero electromagnetic charge. Additional examples can be found here.

The short answer: Peskin & Schroeder assume that the vacuum is invariant under translations and Lorentz transformations (which implies $P^\mu|\Omega\rangle=0$). Alternatively, translation invariance implies $\vec P|\Omega\rangle=0$ in all reference frames, and combining this with invariance under boosts gives $P^0|\Omega\rangle=0$.

The less short answer: Since P&S is perturbative up to chapter 7, one thing you could ask is this: what interactions, when added perturbatively to a given free theory, will automatically lead to the interacting ground state $|\Omega\rangle$ being Lorentz invariant. Perturbation theory assumes that the interacting ground state $|\Omega\rangle$ overlaps with the free vacuum $|0\rangle$, i.e. $\langle\Omega|0\rangle\neq 0$, so let's set $|\Omega\rangle=a|0\rangle+b|\Delta\rangle$ where $\langle \Delta|0\rangle=0$ and $a\neq 0$. By definition of $|0\rangle$ and the 4-momentum for the free theory $P_0^\mu$, we have $P_0^\mu|0\rangle=0$. The 3-momentum operator for the interacting theory is usually the same as that for the free theory (certainly in the absence of derivative couplings), so let's first suppose $\vec P|0\rangle=0$. In order to have $\vec P|\Omega\rangle = \vec p|\Omega\rangle$ with $\vec p\neq 0$ we would need $\vec P|\Delta\rangle \propto |0\rangle$. This is impossible, however, because the orthogonal complement of $|0\rangle$ is preserved by $P_0^\mu$ ($\vec P$ doesn't change particle number). Together with Lorentz invariance, this implies that $P^\mu|\Omega\rangle=0$ when $\vec P=\vec P_0$ perturbatively in the coupling.

One example where $\vec P \neq \vec P_0$ is in an effective field theory for the dynamics of electrons that ignores photons. Here, there is an ambiguity in the ground state from the choice of background photons that could originate from a distant source. If the background field of photons is sufficiently strong, then the 'lowest energy state'* could be a thermalized 'gas' of electron-positron pairs. The 'free' theory for electrons (Dirac fermions) is Lorentz-invariant, but the interacting theory needs to include terms that account for things like finite propagation times (if the gas is sufficiently dilute). Essentially, there is momentum stored in photons that isn't accounted for in the free Fock space of electrons.

In general, the ground state can differ from the (interacting) vacuum state, spontaneously breaking Lorentz symmetry and leading to a preferred frame. Spontaneous symmetry breaking of this type is necessary if $P^\mu|\Omega\rangle\neq 0$, and the preferred frame is the one in which 3-momentum vanishes. Lorentz invariance can be broken in other ways too, e.g. through a vacuum with nonzero electromagnetic charge. Additional examples can be found here.

In Peskin and Schroeder, that $P^\mu|\Omega\rangle=0$ for an interacting theory comes from the assumption that the interacting vacuum has nonzero overlap with the free vacuum, and the fact that the spatial translation operator is the same for free and interacting theories (also that $|\Omega\rangle$ is a momentum eigenstate).

Now theoretically the ground state could differ from the vacuum state, spontaneously breaking Lorentz invariance and leading to a preferred frame (this is necessary if $P^\mu|\Omega\rangle\neq 0$---the preferred frame here is the one in which the 3-momentum vanishes). Lorentz invariance can be broken in other ways too (for example, through a vacuum with nonzero electromagnetic charge. Additional examples can be found here).

The short answer: Peskin & Schroeder assume that the vacuum is invariant under translations and Lorentz transformations (which implies $P^\mu|\Omega\rangle=0$). Alternatively, translation invariance implies $\vec P|\Omega\rangle=0$ in all reference frames, and combining this with invariance under boosts gives $P^0|\Omega\rangle=0$.

The less short answer: Since P&S is perturbative up to chapter 7, one thing you could ask is this: what interactions, when added perturbatively to a given free theory, will automatically lead to the interacting ground state $|\Omega\rangle$ being Lorentz invariant. Perturbation theory assumes that the interacting ground state $|\Omega\rangle$ overlaps with the free vacuum $|0\rangle$, i.e. $\langle\Omega|0\rangle\neq 0$, so let's set $|\Omega\rangle=a|0\rangle+b|\Delta\rangle$ where $\langle \Delta|0\rangle=0$ and $a\neq 0$. By definition of $|0\rangle$ and the 4-momentum for the free theory $P_0^\mu$, we have $P_0^\mu|0\rangle=0$. The 3-momentum operator for the interacting theory is usually the same as that for the free theory (certainly in the absence of derivative couplings), so let's first suppose $\vec P|0\rangle=0$. In order to have $\vec P|\Omega\rangle = \vec p|\Omega\rangle$ with $\vec p\neq 0$ we would need $\vec P|\Delta\rangle \propto |0\rangle$. This is impossible, however, because the orthogonal complement of $|0\rangle$ is preserved by $P_0^\mu$ ($\vec P$ doesn't change particle number). Together with Lorentz invariance, this implies that $P^\mu|\Omega\rangle=0$ when $\vec P=\vec P_0$ perturbatively in the coupling.

One example where $\vec P \neq \vec P_0$ is in an effective field theory for the dynamics of electrons that ignores photons. Here, there is an ambiguity in the ground state from the choice of background photons that could originate from a distant source. If the background field of photons is sufficiently strong, then the 'lowest energy state'* could be a thermalized 'gas' of electron-positron pairs. The 'free' theory for electrons (Dirac fermions) is Lorentz-invariant, but the interacting theory needs to include terms that account for things like finite propagation times (if the gas is sufficiently dilute). Essentially, there is momentum stored in photons that isn't accounted for in the free Fock space of electrons.

In general, the ground state can differ from the (interacting) vacuum state, spontaneously breaking Lorentz symmetry and leading to a preferred frame. Spontaneous symmetry breaking of this type is necessary if $P^\mu|\Omega\rangle\neq 0$, and the preferred frame is the one in which 3-momentum vanishes. Lorentz invariance can be broken in other ways too, e.g. through a vacuum with nonzero electromagnetic charge. Additional examples can be found here.