= Solution
For a small real time displacement $\epsilon$, the <unitary time evolution> operator is $T_t(\epsilon)=1-i\epsilon H+O(\epsilon^2)$. The corresponding <Heisenberg picture> field transformation is conjugation by its inverse:
$$
\begin{aligned}
T_t(\epsilon)^\dagger\phi(t,\mathbf x)T_t(\epsilon)
&=\phi+i\epsilon[H,\phi]+O(\epsilon^2)\\
&=\phi+\epsilon\partial_t\phi+O(\epsilon^2)
=\boxed{\phi(t+\epsilon,\mathbf x)+O(\epsilon^2).}
\end{aligned}
$$
Here <1/a> gives $[H,\phi]=-i\partial_t\phi$.
Likewise let $T_s(\mathbf a)=1+i\mathbf a\cdot\mathbf P+O(|\mathbf a|^2)$, with the paper's sign for its <spatial translation operator>. From <1/b>,
$$
\begin{aligned}
T_s(\mathbf a)^\dagger\phi(t,\mathbf x)T_s(\mathbf a)
&=\phi+i\mathbf a\cdot[\phi,\mathbf P]+O(|\mathbf a|^2)\\
&=\phi+\mathbf a\cdot\nabla\phi+O(|\mathbf a|^2)
=\boxed{\phi(t,\mathbf x+\mathbf a)+O(|\mathbf a|^2).}
\end{aligned}
$$
These are precisely the first-order <Taylor expansions>. Since $H$ and $\mathbf P$ commute in this translation-invariant theory, the finite result is
$$
\boxed{\phi(x)=e^{iP\cdot x}\phi(0)e^{-iP\cdot x},\qquad P\cdot x=Ht-\mathbf P\cdot\mathbf x.}
$$
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