= Solution
The <normal ordering> of both <linear maps> puts every <annihilation operator> to the right, so
$$
H|0\rangle=0,\qquad\mathbf P|0\rangle=\mathbf0.
$$
In particular the stated <Hamiltonian> has already subtracted the free <vacuum energy>. For a <one-particle state>, evaluate the <commutators> using $[AB,C]=A[B,C]+[A,C]B$:
$$
[H,a^\dagger_{\mathbf k}]
=\int_{\mathbf p} E_{\mathbf p}a^\dagger_{\mathbf p}(2\pi)^3\delta^3(\mathbf p-\mathbf k)
=E_{\mathbf k}a^\dagger_{\mathbf k},
\qquad
[\mathbf P,a^\dagger_{\mathbf k}]=\mathbf k\,a^\dagger_{\mathbf k}.
$$
Acting on the <Fock vacuum> therefore gives
$$
\boxed{H|\mathbf k\rangle=E_{\mathbf k}|\mathbf k\rangle,
\qquad\mathbf P|\mathbf k\rangle=\mathbf k|\mathbf k\rangle.}
$$
The extra factor $\sqrt{2E_{\mathbf k}}$ in the chosen <relativistic normalization of a one-particle state> is a scalar multiplier and does not change either <eigenvalue>. Applying the same argument to a <multiparticle quantum state> gives additive total <energy> and <momentum>.
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