Similarly $[\mathbf P,a_{\mathbf p}]=-\mathbf p,a_{\mathbf p}$. Transforming both terms of the field expansion gives
$$
\boxed{e^{-i\mathbf P\cdot\mathbf y}\phi(\mathbf x)
 e^{i\mathbf P\cdot\mathbf y}=\phi(\mathbf x+\mathbf y).}
$$
Therefore the <spatial translation operator> generated by $\mathbf P$ translates the field argument. Equivalently, its action on states translates a wavefunction in the opposite argument convention.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 1 / ii / e / Solution

{title2=$U(\mathbf y)=e^{-i\mathbf P\cdot\mathbf y}$}

The spatial translation operator is the unitary generated by the <momentum operator>. With the convention $U(\mathbf y)=e^{-i\mathbf P\cdot\mathbf y}$, a momentum eigenstate of eigenvalue $\mathbf p$ acquires the phase $e^{-i\mathbf p\cdot\mathbf y}$, while $U(\mathbf y)\phi(\mathbf x)U(\mathbf y)^{-1}=\phi(\mathbf x+\mathbf y)$.

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