One should stress that the single-photon state ##|\vec{p},\lambda \rangle=\hat{a}^{\dagger}(\vec{p},\lambda)|\Omega \rangle## is a generalized momentum eigenstate, i.e., not renormalizable to 1 but only to a ##\delta## distribution in the sense that
$$\langle \vec{p},\lambda|\vec{p}',\lambda' \rangle = (2 \pi)^3 2 |\vec{p}| \delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda,\lambda'}.$$
So, as you said, you need a superposition
$$|\phi,\lambda \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2|\vec{p}|}} \phi(\vec{p}) \hat{a}^{\dagger}(\vec{p},\lambda) |\Omega \rangle,$$
where ##\phi(\vec{p})## is chosen such that
$$\langle \phi,\lambda|\phi,\lambda \rangle= \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3 2|\vec{p}|} |\phi(\vec{p})|^2=1.$$
This is a properly normalized true single-photon state. It's close to a single-mode photon, if ##\phi(\vec{p})## is narrowly peaked around some momentum ##\vec{p}_0##.
This is all quite analogous to classical electrodynamics, where a plane-wave is also not a physical field, because it has infinite energy.
