= Solution
Take $\{\gamma^a,\gamma^b\}=2\eta^{ab}$ and $\gamma^{ab}=\gamma^{[a}\gamma^{b]}=[\gamma^a,\gamma^b]/2$, with unit-weight antisymmetrization. The <Clifford algebra> first gives
$$
\gamma_a\gamma^m\gamma^a=(2-D)\gamma^m,\qquad
\gamma_a\gamma_b\gamma^a=(2-D)\gamma_b.
$$
Expand the two antisymmetrized <gamma matrices> using $\gamma_{ab}=\gamma_a\gamma_b-\eta_{ab}$:
$$
\gamma_{ab}\gamma^m\gamma^{ab}
=\gamma_a\gamma_b\gamma^m\gamma^a\gamma^b-D\gamma^m.
$$
Anticommuting $\gamma^m$ past the next contracted factor gives
$$
\begin{aligned}
\gamma_a\gamma_b\gamma^m\gamma^a\gamma^b
&=-\gamma_a\gamma_b\gamma^a\gamma^m\gamma^b+2\gamma^m\gamma_b\gamma^b\\
&=-(2-D)\gamma_b\gamma^m\gamma^b+2D\gamma^m\\
&=\bigl[2D-(D-2)^2\bigr]\gamma^m.
\end{aligned}
$$
Subtracting the remaining $D\gamma^m$ gives the <gamma-matrix vector sandwich>
$$
\boxed{\alpha=-(D-1)(D-4).}
$$
In particular it vanishes in four dimensions. The result is independent of the choice of Lorentz signature when all indices are contracted consistently. Defining the bivector without its factor $1/2$ would multiply $\alpha$ by four.
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