Covariant derivatives of the frame field, are they the same? There is a exterior covariant derivative
$$
D \alpha_a = d \alpha_a -\omega^b{}_a \wedge \alpha_b\;, \tag 1
$$
and another $D$ in the vierbein postulate
$$
D_\mu e_{I\alpha}=\partial_\mu e_{I\alpha}-\omega_\mu{}^J{}_Ie_{J\alpha}-\Gamma_\mu{}^{\lambda}{}_\alpha e_{I\lambda}=0\;.\tag 2
$$
The question is: Do they the same thing?
 A: You first statement looks like the definition of the torsion form
$$
D{\bf e}^{*a}\equiv  d{\bf e}^{*a}+ {\omega^a}_b \wedge {\bf e}^{*b}= T^a.
$$
The second is a confused aspect of the ordinary covariant derivative and has nothing to do torsion or metricity.
The $e_a^\mu$ and so on are just an array of numbers. Their covariant derivative is just the ordinary partial derivative. The expression you have is therefore not the covaraint derivative of $e^\mu_a$. It is the geometric basis vectors ${\bf e}_a$ or ${\boldsymbol \partial}_\mu$ for the tangent space, or ${\bf e}^{*a}$, or $dx^\mu$  for the cotangent space that need the connection forms ${\omega^b}_{a\mu}dx^\mu$ or ${\Gamma^\lambda}_{\nu\mu}dx^\mu$.
In particular the  "vierbein postulate" is not a postulate. It is an identy that always holds. It is in fact an  appallingly  bad and confusing name for the  formula for converting the  covariant derivative $\nabla_X\equiv X^\mu \nabla_\mu$  from its definition in terms of a  frame-field 
$$
\nabla_X {\bf e}_a= {\bf e}_b{\omega^b}_{a\mu}X^\mu
$$
to its definition in terms of the coordinate basis
$$
\nabla_X {\boldsymbol \partial }_\nu= {\boldsymbol \partial }_\lambda {\Gamma^\lambda}_{\nu\mu} X^\mu.
$$
In each case $\nabla_X$ is the same object. So by expanding
$$
{\bf e}_a =  e^\nu_a {\boldsymbol \partial}_\nu 
$$
and   using   the derivation property (Leibnitz rule) of $\nabla_X$, we   evaluate 
$$
\nabla_X {\bf e}_a =  e_b^\nu {\omega^b}_{a\mu}X^\mu{\boldsymbol \partial}_\nu 
$$
in the equivalent form 
$$
\nabla_X {\bf e}_a  = X^\mu (\partial_ \mu  e^\nu_a){\boldsymbol \partial}_\nu + 
X^\mu e^\nu_a (\nabla_\mu {\boldsymbol \partial}_\nu)\\
=X^\mu (\partial_\mu e_a^\nu+ e^\lambda_a {\Gamma^\nu}_{\lambda \mu}){\boldsymbol \partial}_\nu.
$$
By comparing coefficients of $X^\mu {\boldsymbol \partial}_\nu$, we read off that
$$
\partial_\mu e_a^\nu+ e^\lambda_a {\Gamma^\nu}_{\lambda \mu}- e_b^\lambda {\omega^b}_{a\mu}=0.
$$
Your equation (2) is just an example of this manipulation for the co-frames ${\bf e}^{*a}= e^{*a}_\mu dx^\mu$.  
A: As far as I known the anti-symmetric part of (2) equals to (1).
