I would like to understand an argument in https://link.springer.com/content/pdf/10.1007/BF01393904.pdf, which uses Langlands' result on the constant term of Eisenstein series, but I'm not getting it and I would appreciate any explanation.

On page 429 they define
$$
c(w, \lambda) = \int_{ (w' N_0(\mathbb{A})(w')^{-1} \cap N_0 (\mathbb{A})) \backslash N_0(\mathbb{A}) }
exp(<H_{P_0} ((w')^{-1}n) , \lambda+ \rho_{P_0} >) dn.
$$
Then they say
$$
\int_{N_0(\mathbb{Q}) \backslash N_0(\mathbb{A})} E^G_{P_0}(\lambda, ng) dn
= \sum_{w \in W} c(w, \lambda)
exp( <H_{P_0} (g) , w \lambda + \rho_{P_0} > ) dn.
$$
I think that this statement should follow from the Langlangds' result on page 11 in these slides by Shahidi with $P = P' = P_0$:

https://www.ima.umn.edu/materials/2018-2019/SW11.14-16.18/27644/Shahidi_Abhi.pdf
(or another reference for this is Theorem 6.2.2. (p114) in Shahidi's book "Eisenstein series and autompic L-functions").
But I seem to be missing something and have not been able to deduce the above statement.
For example, in the above $c(w, \lambda)$ is independent of $g$
while in the formula in the link it seems to depend on it...
Any explanation would be appreciated. Thank you.

Some notation: $P_0$ is a minimal $\mathbb{Q}$ parabolic subgroup of $G$ semisimple linear algebraic group over $\mathbb{Q}$. $P_0 = M_0 N_0$ where $M_0$ is a Levi subgroup of $P_0$. And $E^G_{P_0}$ is the Eisenstein series $$ E^G_{P_0}(\lambda, g) = \sum_{\gamma \in P(\mathbb{Q}) \backslash G(\mathbb{Q}) } exp(<\lambda + \rho_{P_0}, H_{P_0} (\gamma g) >). $$

PS I am starting to wonder maybe $c(w,\lambda)$ in the paper (as above) is actually a typo(?) and that it should be replaced by $$ M(w, \lambda)(1)(g) = \int_{ (w' N_0(\mathbb{A})(w')^{-1} \cap N_0 (\mathbb{A})) \backslash N_0(\mathbb{A}) } exp(<H_{P_0} ((w')^{-1}ng) , \lambda+ \rho_{P_0} >) exp(<H_{P_0} (g) , -( w \lambda+ \rho_{P_0}) >) dn $$ as in the formula given in the link above. I am new to the subject and I'm not too sure as maybe I'm missing stuff. Any comments would be appreciated. Thank you.