They start with the time-independent Schrödinger equation:
$$\tag{1}
H|\psi^\alpha\rangle = E^\alpha |\psi^\alpha\rangle.
$$
Then they define $|\psi_0^\alpha\rangle$ to be what they call a "model state", which is an approximation to the true state $|\psi^\alpha\rangle$ but more easily accessible, and is related to the true state by a Møller operator $\Omega$, and to the true energy $E^\alpha$ by $H_{\textrm{eff}}$:
$$\tag{2}
|\psi^\alpha\rangle \equiv \Omega |\psi_0^\alpha\rangle ~~~, ~~~ H_{\textrm{eff}}|\psi_0^\alpha\rangle\equiv E^\alpha |\psi_0^\alpha\rangle.
$$
We can multiply the second part of Eq 2 by $\Omega$ on both sides of the equation, then re-write it using the first part of Eq 2:
\begin{alignat}{2}
\tag{3}
\Omega H_{\textrm{eff}}|\psi_0^\alpha\rangle &= E^\alpha |\psi^\alpha\rangle.
\end{alignat}
Likewise we can simply substitute the first part of Eq 2 into Eq 1 and get:
\begin{alignat}{3}
\tag{4}
H \Omega |\psi_0^\alpha\rangle &= E^\alpha |\psi^\alpha\rangle.
\end{alignat}
Since the right-side of Eqs 3 and 4 are the same, we can equate the left sides too:
\begin{alignat}{3}
\tag{5}
\Omega H_{\textrm{eff}}|\psi_0^\alpha\rangle &= H \Omega |\psi_0^\alpha\rangle.
\end{alignat}
Then they say that since Eq 5 applies for all model states (i.e. all values of $\alpha$) we can replace $|\psi_0^\alpha\rangle$ by $P$:
\begin{alignat}{3}
\tag{6}
\Omega H_{\textrm{eff}}P &= H \Omega P.
\end{alignat}
By partitioning $H=H_0 + V$ we get:
\begin{alignat}{3}
\tag{6}
\Omega H_{\textrm{eff}}P &= H_0 \Omega P + V \Omega P.
\end{alignat}
What happens next is not explained in the text, but they get:
\begin{alignat}{3}
\tag{7}
H_{\textrm{eff}} &= PH_0 P + PV \Omega P.
\end{alignat}
To get the last term, I would left-multiply all terms by $P$, but something else must have happened for the two other terms to be the way they are. Anyway, they make the second term in Eq. 7 to be defined as $W$ and remove the $H_{\textrm{eff}}$ to get the equation at the beginning of your question. They cite [113,117,124] but with your Google Books link, I can't see pages 61 to 405 and it seems the bibliography is at the end (in those missing pages). Do you happen to know what [113,117,124] are?
Anyway, they finally say that in the case where all model states have the same energy $E_0$, we get the original Bloch equation that they say was derived in the late 1950s by the French nuclear physicist Claude Bloch, (not the much more famous Bloch equation first presented in 1948 by the Swiss-American physicist Felix Bloch and his PhD student Roald Wangsness):
$$\tag{8}
\left(E_0 - H_0 \right)\Omega P = V \Omega P - \Omega W.
$$
Edit: After a lot more searching I have found that the author of that book explained this same derivation in a lot more detail in this 1974 paper. Pay attention to what he says about $P$ and $\Omega$ not being "true" inverses of each other, and what happens when you combine them. It is also quite amazing that the notation is almost the same in the original 1958 paper of Claude Bloch. The paper is in French, but it's not too hard to follow the equations (he shows quite a lot of steps!).