I don't feel that I have a mastery of the equation in your question, but I have some things to offer that can at least help.
In QM, if we resolve the single-particle identity in the momentum basis we see
$$1 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} |\vec{p} \rangle \langle \vec{p}| d^3p $$
We want something similar in QFT. But the hilbert space in QFT includes states with 0 particles, or 1 particle, or 2 particles, etc. So we need more terms here, which account for states like $|\Omega \rangle$ or $|p_1 p_2 \rangle$. We also want Lorentz covariance, so we need to use a different measure. The general idea is to write something like
$$1 = |\Omega \rangle \langle \Omega | + \int_{-\infty}^{\infty} |p \rangle \langle p| \frac{d^3p}{(2\pi)^3 2E_p} + (\text{2-particle identity}) + ... \tag{1}$$
Then we have covered all states. Because otherwise, the identity acting on for example $|\Omega \rangle$ or $|p_1 p_2 \rangle$ would just give zero! Which is wrong.
The equation you wrote is in the spirit of Eq. (1) above. But, there are deeper reasons why there are more complications than what we have established so far. Namely, an identity operator resolution like (1) uses states $|p \rangle $ constructed from creation/annihilation operators that come from the solution to the equation of motion of the free theory, without interacting terms:
$$a^\dagger(p)|\Omega \rangle = |p \rangle$$
but in an interacting theory, we may not have the momentum eigenstates from the free theory. Generally, the hilbert space of the interacting theory is not the same. For this reason, the use of the label $\lambda_p$ leaves more leeway. Taking a look at the equation you linked,
$$
\mathbf{1}=|\Omega\rangle\langle\Omega|+\sum_{\lambda} \int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{2 E_{\mathbf{p}}(\lambda)}\left|\lambda_{\mathbf{p}}\right\rangle\left\langle\lambda_{\mathbf{p}}\right|
$$
The integral here is over the total momentum of all particles combined, and $|\lambda_p \rangle$ is a multi-particle state with total momentum $\vec{p}$. The "mysterious" part to me is the sum over $\lambda$, and even the professor with whom I took QFT could not tell me exactly what this sum consisted of. But in principle, it should sum over all states - of any particle number - which have total momentum $\vec{p}$, and for which the total four-momentum operator has eigenvalue
$$P^2 |\lambda_p \rangle = m_\lambda^2 |\lambda_p \rangle$$
Such that, then, $E_p$ is defined as
$$E_p = \sqrt{m_\lambda^2 + \vec{p}^2}$$
For that reason $m_\lambda$ is not the mass of any individual particle but just a defined label for this state in general.
Best Answer
You can't - it is just there by analogy. 7.2 is the version of 7.1 for the hilbert space in QFT
$|\lambda_0 \rangle$ includes all eigenstates of $H$ and $P$, namely the total energy and momentum of all particles in the system. It does not include $|\Omega \rangle \langle \Omega |$ which is pulled out of the sum separately for convenience and to make it clear that it is in fact included.
For the $|\lambda_0 \rangle$ to be "relativistically normalized", the entire identity acting on some $|\lambda'_{p'}\rangle$ should, through a delta function and proper cancellation of the factor including the energy in the integrand, give back the same $|\lambda'_{p'}\rangle$. For this to happen, we need the normalization of $|\lambda_{p}\rangle$ to depend on the energy. The reason it would be called relativistically normalized is that it cancels out with the factor which makes the integrand relativistically covariant, namely
$$\frac{d^3 p}{(2 \pi)^3 2E_p(\lambda)}$$