This is exercise $7.22$ in Supergravity by Freedman and Van Proeyen, but I did not understand it and would appreciate if you clear it out.
Given the below, I still don't get how, if we define the last equation I wrote to hold, we will get $T'^a=\Lambda^{-1a}_{\;b}T^b$ if I combine the transformations below in $de^a+ω^a_b∧e^b$. It is not working with me.
If the one-form is
$$e^a = e_{\mu}^adx^{\mu}$$
then the two-form will be
$$de^a = \frac{1}{2}(\partial_{\mu}e^a_{\nu} - \partial_{\nu}e^a_{\mu})dx^{\mu}\wedge dx^{\nu}.$$
This doesn't transform well under Lorentz transformation because the second term spoils the transformation:
$$de'^a = d(\Lambda^{-1a}_{\;b}e^b)=\Lambda^{-1a}_{\;b}de^b+d\Lambda^{-1a}_{\;b}\wedge e^b.$$
To cancel this term, we add the contribution from the two-form involving the spin connection and consider
$$de^a+\omega^a_{\;b}\wedge e^b=T^a.$$
So $T^a$ transforms as
$$T'^a=\Lambda^{-1a}_{\;b}T^b$$ if we define that $\omega^a_{\;b}$ to transform as
$$ \omega^{'a}_b=\Lambda^{-1a}\hspace{.5mm}_c d\Lambda^c\hspace{.5mm}_b +\Lambda^{-1a}\hspace{.5mm}_c\omega^c\hspace{.5mm}_d\Lambda^d\hspace{.5mm}_b.$$
Note that
\begin{align*} T'^{a} &= de'^a + \omega'^a_b\wedge e'^b\\ &= d((\Lambda^{-1})^a_b e^b) + ((\Lambda^{-1})^a_cd\Lambda^c_b + (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b)\wedge e'^b\\ &= (\Lambda^{-1})^a_bde^b + d(\Lambda^{-1})^a_b\wedge e^b + (\Lambda^{-1})^a_cd\Lambda^c_b\wedge e'^b + (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b\wedge e'^b\\ &= (\Lambda^{-1})^a_bde^b + d(\Lambda^{-1})^a_b\wedge e^b + \underbrace{(\Lambda^{-1})^a_cd\Lambda^c_b\wedge(\Lambda^{-1})^b_de^d}_{(1)} + \underbrace{(\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b\wedge(\Lambda^{-1})^b_fe^f}_{(2)}. \end{align*}
Now let's simplify terms $(1)$ and $(2)$.
$(1)$
As $\delta^a_b = (\Lambda\Lambda^{-1})^a_b = \Lambda^a_c(\Lambda^{-1})^c_b$, we see that $0 = d(\Lambda^a_c(\Lambda^{-1})^c_b) = d\Lambda^a_c(\Lambda^{-1})^c_b + \Lambda^a_cd(\Lambda^{-1})^c_b$. Therefore
\begin{align*} (\Lambda^{-1})^a_cd\Lambda^c_b\wedge(\Lambda^{-1})^b_de^d &= (\Lambda^{-1})^a_cd\Lambda^c_b(\Lambda^{-1})^b_d\wedge e^d\\ &= -(\Lambda^{-1})^a_c\Lambda^c_bd(\Lambda^{-1})^b_d\wedge e^d\\ &= -(\Lambda^{-1}\Lambda)^a_bd(\Lambda^{-1})^b_d\wedge e^d\\ &= - \delta^a_bd(\Lambda^{-1})^b_d\wedge e^d\\ &= - d(\Lambda^{-1})^a_d\wedge e^d. \end{align*}
Swapping the index $d$ with the index $b$, we see that term $(1)$ becomes $- d(\Lambda^{-1})^a_b\wedge e^b$.
$(2)$
We have
\begin{align*} (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b\wedge(\Lambda^{-1})^b_fe^f & = (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b(\Lambda^{-1})^b_f\wedge e^f\\ &= (\Lambda^{-1})^a_c\omega^c_d(\Lambda\Lambda^{-1})^d_f\wedge e^f\\ &= (\Lambda^{-1})^a_c\omega^c_d\delta^d_f\wedge e^f\\ &= (\Lambda^{-1})^a_c\omega^c_d\wedge e^d. \end{align*}
Swapping the index $c$ with the index $b$, we see that term $(2)$ becomes $(\Lambda^{-1})^a_b\omega^b_d\wedge e^d$.
Combining these results, we see that
\begin{align*} T'^{a} &= (\Lambda^{-1})^a_bde^b + d(\Lambda^{-1})^a_b\wedge e^b + (\Lambda^{-1})^a_cd\Lambda^c_b\wedge(\Lambda^{-1})^b_de^d + (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b\wedge(\Lambda^{-1})^b_fe^f.\\ &= (\Lambda^{-1})^a_bde^b + d(\Lambda^{-1})^a_b\wedge e^b - d(\Lambda^{-1})^a_b\wedge e^b + (\Lambda^{-1})^a_b\omega^b_d\wedge e^d\\ &= (\Lambda^{-1})^a_bde^b + (\Lambda^{-1})^a_b\omega^b_d\wedge e^d\\ &= (\Lambda^{-1})^a_b(de^b + \omega^b_d\wedge e^d)\\ &= (\Lambda^{-1})^a_bT^b. \end{align*}