= Solution
For the spatial <momenta>, integration by parts gives
$$
\frac{\delta P_i}{\delta\pi_a}=\partial_i\phi_a,\qquad\frac{\delta P_i}{\delta\phi_a}=-\partial_i\pi_a.
$$
Their <Poisson bracket> is therefore
$$
\begin{aligned}
\{P_i,P_j\}_{\rm PB}&=\int d^3x\,[-\partial_i\pi_a\partial_j\phi_a+\partial_i\phi_a\partial_j\pi_a]\\
&=\int d^3x\,\pi_a(\partial_i\partial_j\phi_a-\partial_j\partial_i\phi_a)=0.
\end{aligned}
$$
For the mixed case, $P_0=H$ and the spatial generator acts on both canonical fields by differentiation: $\{\phi_a,P_i\}=\partial_i\phi_a$, $\{\pi_a,P_i\}=\partial_i\pi_a$. Hence, for a translation-invariant local <Hamiltonian density>,
$$
\{H,P_i\}_{\rm PB}=\int d^3x\,\partial_i\mathcal H=0.
$$
Finally $\{H,H\}_{\rm PB}=0$ by antisymmetry. Together these prove
$$
\boxed{\{P_\mu,P_\nu\}_{\rm PB}=0.}
$$
The proof requires commuting derivatives and no surviving boundary terms. It expresses the Abelian translation subgroup of <Poincaré symmetry> at the level of its canonical generators.
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