= Solution
Write $d\Pi_p=d^3p/[(2\pi)^3 2\omega_p]$, $p^\mu=(\omega_p,\mathbf p)$ and $\omega_p=\sqrt{\mathbf p^2+m^2}$. The <Fourier expansion> in the source is compatible with the oscillator <canonical commutation relations>
$$
[a_{\mathbf p},a_{\mathbf q}^\dagger]=(2\pi)^3 2\omega_p\delta^{(3)}(\mathbf p-\mathbf q),\qquad[a,a]=[a^\dagger,a^\dagger]=0.
$$
For the free real <scalar field>, $\pi=\dot\phi$ and
$$
P_0=\frac12\int d^3x\,[\pi^2+(\nabla\phi)^2+m^2\phi^2],\qquad P_i=\int d^3x\,\pi\partial_i\phi.
$$
Substitution of the <Fourier expansion> produces spatial <Dirac delta functions>. Terms with two <annihilation operators> or two <creation operators> cancel: in $P_0$ their coefficient is $-\omega_p^2+\mathbf p^2+m^2=0$, and in $P_i$ their integral is odd under $\mathbf p\mapsto-\mathbf p$. The two mixed terms combine, with <normal ordering> removing the vacuum constant, to give
$$
\boxed{\hat P_\mu=\int d\Pi_p\,p_\mu a_{\mathbf p}^\dagger a_{\mathbf p}.}
$$
In particular, $P_i$ contains the lower component $p_i=-\mathbf p_i$ in this <Minkowski metric> convention.
To check commutativity without manipulating a square of a <Dirac delta function>, define $N_f=\int d\Pi_p f(p)a_p^\dagger a_p$. The oscillator algebra gives $[N_f,a_q]=-f(q)a_q$ and $[N_f,a_q^\dagger]=f(q)a_q^\dagger$. Thus $[N_f,a_q^\dagger a_q]=0$ and $[N_f,N_g]=0$. Taking $f=p_\mu$, $g=p_\nu$ proves
$$
\boxed{[\hat P_\mu,\hat P_\nu]=0.}
$$
These identities hold on the usual dense domain of finite-particle states, or can first be proved in a finite box and then continued to the continuum.
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