2D Ising model and FK-percolation

Consider the 2D Ising model on the finite lattice $Lambda$ with $+$ boundary conditions, i.e., all spins outside of $Lambda$ are $=+1$. Let $mathscr{E}_Lambda^b$ denote the edges in $Lambda$ and the edges connecting $Lambda,Lambda^c$ so that the Hamiltonian is given by
$$
H = H_{Lambda;beta,0}^+ (sigma) = -beta sum_{klin mathscr{E}_Lambda^b}sigma_ksigma_l
$$
By writing
$$
e^{beta sigma_k sigma_l}=e^beta((1-p)+p1_{sigma_k=sigma_l}), quad p=1-e^{-2beta}
$$
We can deduce that the partition function (as done in Velenik’s Stat Mech of Lattice Systems, Chap 3.10.6) is given by
$$
Z_{Lambda}^+ = e^{beta |mathscr{E}_Lambda^b} sum_{Esubset mathscr{E}_Lambda^b} p^{|E|}(1-p)^{|mathscr{E}_Lambda^b backslash E|} sum_{omegain Omega_Lambda^+} prod_{klin E} 1_{sigma_k(omega)=sigma_l(omega)}
$$
where $Omega_Lambda^+$ denotes the possible spin configurations on $Lambda$ with all spins outside of $Lambda$ fixed to be $=+1$. In the next step, Velenik claims that
$$
Z_{Lambda}^+ = e^{beta |mathscr{E}_Lambda^b} sum_{Esubset mathscr{E}_Lambda^b} p^{|E|}(1-p)^{|mathscr{E}_Lambda^b backslash E|} 2^{N_Lambda^w(E)-1}
$$
where $N_Lambda^w(E)$ is the number of connected components of the graph $(mathbb{Z}^d, Ecup mathscr{E}_{Lambda^c}$).

Question. Shouldn’t it be
$$
sum_{omega in Omega_Lambda^+} prod_{klin E} 1_{sigma_k(omega)=sigma_l(omega)} = 2^{N_Lambda^w(E)-1} 2^{|Lambda backslash V_E|}
$$
where $V_E$ is the set of vertices of $E$, since the spins on $Lambda backslash V_E$ are free to change? If so, why would the 2D Ising model correspond to the FK-percolation process now that we have an extra $2^{|Lambda backslash V_E|}$ term?