Here's my attempt:
WLOG we assume that $p$ is even, so each of the binary vectors $b^A, b^B$ associated with the models (and with assessment of configural equivalence) has exactly $\frac{p}{2}$ 0's and $\frac{p}{2}$ 1's. If we take binary vector $b^A$ as is, its corresponding ranks would be $\frac{p+2}{4}$ for the 0's and $\frac{3p+2}{4}$ for the 1's.
However, if we add some insignificant "noise" component $\tilde{b}^A_i=b^A_i+\eta^A_i,\quad \eta^A_i\sim\mathcal{N}\left((r^A_i-\frac{p+1}{2})\cdot 10^{-6}, 10^{-12}\right)$ then the ranks are integers again. The 0's will get the ranks $1,…,\frac{p}{2}$ and the 1's will get ranks $\frac{p+2}{2},…,p$. We can safely say that $\tilde{b}^A_i\approx b^A_i$ but much more important - due to the construction of $\eta_i$ we get that the rank vector $R\left[\tilde{b}^A\right]$ of $\tilde{b}^A$ is identical to $r^A$ the original rank vector of the mean absolute SHAP values. That is, $r_s\left(\tilde{b}^A, \tilde{b}^B\right)=r_s \left(s^A, s^B\right)$.
First, we would like to get an approximation for $\rho\left(\tilde{b}^A, \tilde{b}^B\right)$:
$$\begin{aligned} E\left[\tilde{b}^A\right] = \frac{1}{p}\sum_{i=1}^p{\tilde{b}^A_i} = \frac{1}{p}\left(\sum_{i=1}^p{b^A_i}+\sum_{i=1}^p{E\left[\eta^A_i\right]}\right) = \bar{b}^A+\frac{10^{-6}}{p}\sum_{i=1}^p{\left(r^A_i-\frac{p+1}{2}\right)} = \\\bar{b}^A+\frac{10^{-6}}{p}\left(\frac{p(p+1)}{2}-\frac{p(p+1)}{2}\right) = \bar{b}^A = \frac{1}{2} \end{aligned}$$
$$\begin{aligned}\left( \tilde{b}^A_i - \bar{\tilde{b}}^A \right)^2=\left( b^A_i + \eta^A_i - \bar{b}^A \right)^2 = \\ \left( b^A_i + \eta^A_i \right)^2-2\bar{b}^A\left( b^A_i + \eta^A_i \right)+\left(\bar{b}^A\right)^2 = \\ \left( b^A_i\right)^2 + 2b^A_i\eta^A_i + \left( \eta^A_i\right)^2 - 2\bar{b}^Ab^A_i - 2\bar{b}^A\eta^A_i + \left(\bar{b}^A\right)^2 = \\ \left(b^A_i - \bar{b}^A\right)^2 + 2\left(b^A_i - \bar{b}^A\right) \eta^A_i + \left( \eta^A_i\right)^2\end{aligned}$$
$$\tilde{s}^A = \sqrt{\sum_{i=1}^p{\left( \tilde{b}^A_i - \bar{\tilde{b}}^A \right)^2}} = \sqrt{ \underset{(s^A)^2}{\underbrace{\sum_{i=1}^p{\left(b^A_i - \bar{b}^A\right)^2}}} + \underset{\sim10^{-6}}{\underbrace{\sum_{i=1}^p{2\left(b^A_i - \bar{b}^A\right) \eta^A_i}}} + \underset{\sim10^{-12}}{\underbrace{\sum_{i=1}^p{\left( \eta^A_i\right)^2}}}} \approx s^A$$
$$\begin{aligned}
\left(\tilde{b}^A_i - \bar{\tilde{b}}^A\right)\left(\tilde{b}^B_i - \bar{\tilde{b}}^B\right) = \left(\tilde{b}^A_i - \bar{b}^A\right)\left(\tilde{b}^B_i - \bar{b}^B\right) =\\ \left(b^A_i - \bar{b}^A + \underset{\sim10^{-6}}{\underbrace{\eta^A_i}}\right)\left(\tilde{b}^B_i - \bar{b}^B\right) \approx \left(b^A_i - \bar{b}^A\right)\left(\tilde{b}^B_i - \bar{b}^B\right)\\ \approx \left(b^A_i - \bar{b}^A\right)\left(b^B_i - \bar{b}^B\right)
\end{aligned}$$
Putting it all together, we get
$$\rho\left(\tilde{b}^A, \tilde{b}^B\right) = \frac{\sum_{i=1}^p{\left(\tilde{b}^A_i - \bar{\tilde{b}}^A\right)\left(\tilde{b}^B_i - \bar{\tilde{b}}^B\right)}}{\tilde{s}^A\cdot\tilde{s}^B} \approx \frac{\sum_{i=1}^p{\left(b^A_i - \bar{b}^A\right)\left(b^B_i - \bar{b}^B\right)}}{ s^A\cdot s^B} = \rho\left(b^A,b^B\right).$$
With that in hand, we can move on. When looking at vectors $\tilde{b}^A, \tilde{b}^B$ we get another important property: the pair $\tilde{b}^A, \tilde{b}^B$ has a bivariate distribution, due to $\eta_i$ being normal . This means we can use the Spearman-to-Pearson conversion formula, $\rho=2\sin\left(\frac{\pi}{6}\cdot r_s\right)$.
Next, we note that the function $\sin\left(\frac{\pi}{6}\cdot x\right)$ is monotone in $x\in[-1,1]$. If $r_s > r^*$ then $\frac{\pi}{6}\cdot r_s > \frac{\pi}{6}\cdot r^*$. As both $r_s,r^*$ are in $[-1,1]$, we can apply the sine to both sides, getting $\sin\left(\frac{\pi}{6}\cdot r_s\right) > \sin\left(\frac{\pi}{6}\cdot r^*\right)$. Multiplying both sides by 2, we get $2\sin\left(\frac{\pi}{6}\cdot r_s\right) > 2\sin\left(\frac{\pi}{6}\cdot r^*\right)$, eventually $\rho > 2\sin\left(\frac{\pi}{6}\cdot r^*\right)$.
Finally, if $r_s\left(r^A, r^B\right) > r^*$ then also $r_s\left(\tilde{b}^A, \tilde{b}^B\right) > r^*$, meaning $\rho\left(\tilde{b}^A, \tilde{b}^B\right) > 2\sin\left(\frac{\pi}{6}\cdot r^*\right)$. Due to $\rho\left(\tilde{b}^A, \tilde{b}^B\right) \approx \rho\left(b^A,b^B\right)$, we get the boundary $\rho\left(b^A, b^B\right) > 2\sin\left(\frac{\pi}{6}\cdot r^*\right)$. $\blacksquare$
