= Solution
Both a <particle> and its <antiparticle> carry the <momentum> of their mode label. Thus the <normal-ordered> total <momentum operator> is
$$
\boxed{\mathbf P=\int_{\mathbf p}\mathbf p\left(a_{\mathbf p}^\dagger a_{\mathbf p}+b_{\mathbf p}^\dagger b_{\mathbf p}\right).}
$$
The <canonical commutation relations> give
$$
[a_{\mathbf p},\mathbf P]=\mathbf p\,a_{\mathbf p},\qquad
[b_{-\mathbf p}^\dagger,\mathbf P]=-(-\mathbf p)b_{-\mathbf p}^\dagger=\mathbf p\,b_{-\mathbf p}^\dagger.
$$
The minus sign in the <antiparticle> mode label is essential: both terms in the printed <mode expansion of a free field> contain $e^{i\mathbf p\cdot\mathbf x}$. Hence
$$
[\phi,\mathbf P]=\int_{\mathbf p}\frac{\mathbf p}{\sqrt{2E_p}}\left(b_{-\mathbf p}^\dagger e^{iE_pt}+a_{\mathbf p}e^{-iE_pt}\right)e^{i\mathbf p\cdot\mathbf x}
=\boxed{-i\nabla\phi.}
$$
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