How to vary fermion action in the index-free Clifford notation with respect to spin-connection? In ref (1), it is claimed that the Dirac action (2.30) 
\begin{equation}
S_D \sim \int ( \overline \psi \star eee D \psi + \overline {D\psi} \star eee  \psi)
\end{equation}
becomes
\begin{equation}
\delta_\omega S_D \sim \int   - \{\psi\overline \psi, \star eee\} \;\delta \omega
\end{equation}
where varying the action with respect to the spin connection $\omega$. I don't understand why we must rearrange the terms in action this way. 
References
(1) A.C.Randono, 'In Search of Quantum de Sitter Space: Generalizing the
Kodama State'
link
 A: \begin{equation}
\delta_\omega S_D \sim \int Tr ( \overline \psi \star eee (-\delta \omega) \psi + (-\delta \omega \overline {\psi}) \star eee  \psi)\\
=\int Tr ( -\overline \psi \star eee  \psi - \overline {\psi} \star eee  \psi)\delta \omega\\
\end{equation}
By using $Tr(ABC)=Tr(CAB)=Tr(BCA)$ we can write
$$
S_D \sim \int Tr ( - \star eee  \psi \overline \psi- \psi\overline {\psi} \star eee  )\delta \omega\\
$$
This is what you want.
A: \begin{equation}
 D_\mu \psi^A = \partial_\mu \psi^A - \frac i 2 \omega_\mu{}^{IJ}(\mathcal J_{IJ})^A{}_B \, \psi^B\;,
\end{equation}
where $\mathcal J_{IJ}\equiv - \frac i 4 [\gamma_I, \gamma_J]$. Or we may simply write
\begin{eqnarray}
 D_\mu \psi^A &=&\partial_\mu\psi^A  - \frac 1 8 \omega_\mu{}^{IJ}[\gamma_I,\gamma_J]^A{}_B\, \psi^B\;,\\
    &=& \partial_\mu\psi^A  - \frac 1 4 \omega_\mu{}^{IJ}(\gamma_I,\gamma_J)^A{}_B\, \psi^B\;.
\end{eqnarray}
\begin{eqnarray}
 \delta_\omega(D_\mu \psi^A ) &=&- \frac 1 4 \delta(\omega_\mu{}^{IJ})(\gamma_I,\gamma_J)^A{}_B\, \psi^B\;,\\
    &\equiv& - \delta(\omega_\mu{}^{IJ})^A{}_B\, \psi^B\;,
\end{eqnarray}
also, since we use the real representation of $\gamma$'s (In this thesis it use $\{\gamma_I,\gamma_J\}=+2\eta_{IJ}$ with $\eta=diag(-1,1,1,1)$ )
\begin{equation}
 \delta_\omega(\overline{D_\mu \psi_A} ) = - \delta(\omega_\mu{}^{IJ})^B{}_A\,\overline{ \psi_B}\;.
\end{equation}
In shorthand notation we write
\begin{eqnarray}
 \delta_\omega \, D\psi &=& -\delta\omega \,\psi \;,\\
  \delta_\omega \,\overline {D\psi} &=& - \overline {\psi}\,\delta\omega \;
\end{eqnarray}
Note that to see $\omega$ in Clifford basis we try
\begin{eqnarray}
 \omega= \frac 1 4 \omega_{IJ}\gamma^I \gamma^J ,\; \omega \wedge \omega &=& \frac 1 {16} \omega_{IJ} \wedge \omega_{KL} \gamma^I < \gamma^J,\gamma^K> \gamma^L\;,\\
 &=& \frac 1 {4} \omega_{IK} \wedge \omega^K{}_{L} \gamma^I  \gamma^L\;.\\
 F = \mathrm{d} \omega+ \omega \wedge \omega &=& \frac 1 4 \mathrm{d}\omega_{IJ}\gamma^I \gamma^J + \frac 1 {4} \omega_{IK} \wedge \omega^K{}_{L} \gamma^I  \gamma^L\;,\\
 &=&\frac 1 4 \Big[   \mathrm{d}\omega_{IJ}+  \omega_{IK} \wedge \omega^K{}_{J} \Big] \gamma^I \gamma^J \equiv \frac 1 4 F_{IJ} \gamma^I \gamma^J\;.
\end{eqnarray}
Which consistence to $F$ in the thesis.
\begin{eqnarray}
 S &\sim& \int\Big( \overline \psi_A (\star e e e)^A{}_B (D\psi)^B + (\overline{D\psi})_A (\star eee)^A{}_B \psi^B  \Big)\;,\\
 \delta_\omega S &\sim& \int \Big( \overline \psi_A (\star e e e)^A{}_B (-\delta \omega\,\psi)^B + (-\overline{\psi}\delta \omega)_A (\star eee)^A{}_B \psi^B  \Big)\;,\\
 &=&\int - \overline \psi_A (\star e e e)^A{}_B \delta \omega^B{}_C\,\psi^C -\overline{\psi}_C\delta \omega^C{}_A (\star eee)^A{}_B \psi^B \;,\\
 &=& \int - \psi^C \overline \psi_A (\star e e e)^A{}_B \delta \omega^B{}_C\,- (\star eee)^A{}_B \psi^B \overline{\psi}_C\delta \omega^C{}_A\;,\\
 (&\equiv& \int - \psi^C \otimes \overline \psi_A (\star e e e)^A{}_B \delta \omega^B{}_C\,- (\star eee)^A{}_B \psi^B \otimes\overline{\psi}_C\delta \omega^C{}_A)\;,\\
 &=&   \int - \psi \overline \psi (\star e e e) \delta \omega\,- (\star eee)\psi \overline{\psi}\delta \omega\;,\\
 &=&   -\int  \Big\{\psi \overline \psi ,(\star e e e) \Big\}\delta \omega\,
\end{eqnarray}
