s}
\sum_{KkI}
\int_{b\Omega}D^\alpha \phi_I \overline{D^\alpha \psi_K}
\pd{\varrho}{\bar z_k} .$$ By integrating by parts we can move tangential derivatives from $\phi$ to $\psi$, so only the $s$ normal derivatives on $\phi$ may cause trouble.

We decompose the standard derivatives in the coordinate directions into their normal and tangential components: $$D_j = Y_j +\nu_j N,$$ where $N$ is the normal derivative, and $Y_j$ are tangential vector fields. Then $$D^\alpha=(Y_{\alpha_{p_1}}+\nu_{\alpha_{p_1}}N)\cdots
(Y_{\alpha_{p_s}}+\nu_{\alpha_{p_s}}N).$$ Notice that, since $\sum_j \nu_j^2 \equiv 1$ and $N=\sum_j \nu_j D_j$, we have that $\sum_j \nu_j Y_j =0$. Therefore, when considering $s$ normal derivatives on $\phi_I$, we have $$\begin{aligned}
\lefteqn{\sum_{|\alpha|=s}\CP \sum_{KkI} } \\
& \biggl [ \e{K}{kI} \int_{b\Omega} 
(\nu_{\alpha_{p_1}}N)\cdots
(\nu
_{\alpha_{p_s}}N)\phi_I
\overline{(Y_{\alpha_{p_1}}+\nu_{\alpha_{p_1}}N)\cdots 
(Y_{\alpha_{p_s}}+\nu_{\alpha_{p_s}}N) \psi_K} 
\pd{\varrho}{\bar z_k} \biggr ] \\
& = \sum_{|\alpha|=s} \CP \sum_{KkI} \e{K}{kI} \int_{b\Omega} 
(\nu_{\alpha_{p_1}})^2 \cdots (\nu_{\alpha_{p_s}})^2 
\bigl( N^s \phi_I\bigr) \overline{\bigl(N^s \psi_K)}
    \pd{\varrho}{\bar z_k} \\
& = \bigl( \sum_{|\alpha|=s}\CP (\nu_{\alpha_{p_1}})^2 \cdots 
(\nu_{\alpha_{p_s}})^2 \bigr)\sum_I \int_{b\Omega} 
(N^s \phi_I) \overline{\bigl
( \sum_{Kk} \e{K}{kI} N^s (\psi_K
\pd{\varrho}{z_k} )\bigr) } . \end{aligned}$$ Now, if $\psi\in C^\infty_{(0,q+1)} (\overline{\Omega})$ and $$0=\sum_{Kk} \e{K}{kI} N^s (\psi_K
\pd{\varrho}{z_k}) = N^s (\psi\llcorner \dbar\varrho)_I$$ on $b\Omega$ for all $I$, then clearly $\psi\in\dom\dbar^*$.

On the other hand, suppose that $N^s (\psi\llcorner\dbar\varrho)_I \neq 0$ on $b\Omega$ for a certain $I$. We may assume that 