= Solution
Assume $m_\psi>0$, as required to have a <rest frame>. Choose the <Lorentz boost> $\Lambda$ for which $\Lambda p=(m_\psi,\mathbf0)$, and write its <unitary operator> as $U$. The stipulated <one-particle state> normalization gives $U|\psi,\mathbf p\rangle=|\psi,\mathbf0\rangle$ without a momentum-dependent prefactor. A <Lorentz scalar> at the origin satisfies $U\phi(0)U^\dagger=\phi(0)$, and the <vacuum state> satisfies $U|\Omega\rangle=|\Omega\rangle$. Inserting $U^\dagger U$ therefore proves
$$
\boxed{\langle\Omega|\phi(0)|\psi,\mathbf p\rangle=\langle\Omega|\phi(0)|\psi,\mathbf0\rangle.}
$$
With a noncovariant normalization of <momentum eigenstates>, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.
Let $q^\mu=(E_{\mathbf q}^{\chi},\mathbf q)$ denote the <four-momentum> of the other <energy eigenstate>. Translation covariance first gives
$$
\langle\chi,\mathbf q|\phi(x)|\psi,\mathbf p\rangle
=e^{i(q-p)\cdot x}\langle\chi,\mathbf q|\phi(0)|\psi,\mathbf p\rangle
=e^{i(q-p)\cdot x}\langle U\chi,\mathbf q|\phi(0)|\psi,\mathbf0\rangle.
$$
Thus a suitable ket, including the phase whose complex conjugate appears in its bra, is
$$
\boxed{|\Psi\rangle=e^{-i(q-p)\cdot x}U|\chi,\mathbf q\rangle.}
$$
Here $\langle U\chi,\mathbf q|$ means $(U|\chi,\mathbf q\rangle)^\dagger$. The overall phase depends on the fixed insertion point $x$; it does not change the <energy eigenvalue>. It cannot simply be dropped if the equality of complex <matrix elements> is to hold exactly.
The transformed state has <four-momentum> $q'=\Lambda q$. <Lorentz invariance> of the <Minkowski inner product> gives
$$
p\cdot q=(\Lambda p)\cdot(\Lambda q)=m_\psi q'^0,
\qquad
\boxed{E^\Psi=\frac{p\cdot q}{m_\psi}=\frac{E_{\mathbf p}E_{\mathbf q}^{\chi}-\mathbf p\cdot\mathbf q}{m_\psi}.}
$$
There is no boost to a <rest frame> for a nonzero massless <four-momentum>, so the rest-state statements require the positive-mass assumption.
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