The problem is sloppy (but convenient) notation. The objects, \begin{equation} \left| k \right] ^a , \quad \left| p \right\rangle ^{ \dot{a} } \end{equation}
are two component spinors while $ \gamma_\mu$ is a 4x4 matrix. So its even clear what the brakets mean. When we write the braket, \begin{equation} \left[ k | \gamma ^\mu | p \right\rangle \end{equation} what we really mean is that we pick out the $ \sigma ^\mu $ matrix in the $ \gamma ^\mu $ with the correct index structure. For example, \begin{equation} \left[ k \right| ^a \left( \begin{array}{cc} 0 & \sigma ^\mu _{ a \dot{a} } \\ \bar{\sigma} ^\mu _{ \dot{a} a } & 0 \end{array} \right) \left| p \right\rangle^ {\dot{a}} = \left[ k \right| ^a \sigma ^{ \mu } _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \end{equation} with similar results for the other brakets.

Now getting back to your question. The $ \gamma ^\mu $ matrix is not invariant under complex conjugation: \begin{equation} \left( \gamma ^\mu \right) ^\dagger = \gamma _0 \gamma ^\mu \gamma _0 = \left( \begin{array}{cc} 0 & \bar{\sigma} ^\mu \\ \sigma ^\mu & 0 \end{array} \right) \end{equation} so all complex conjugation does is switch the positions of the $ \sigma ^\mu $ and $ \bar{\sigma} ^\mu $ matrices. Therefore, we can just omit the complex conjugation if we remember the ``pick Pauli matrix with the correct index structure'' prescription.

Then we have, \begin{equation} \left( \left[ k \right| ^a ( \gamma ^\mu ) _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \right) ^\ast = \left[ p \right| \gamma ^\mu \left| k \right\rangle \end{equation} where it is understood that now we are picking out the $ \bar{\sigma} ^\mu $ matrix instead of $ \sigma ^\mu $ in $ \gamma ^\mu $.

The problem is sloppy (but convenient) notation. The objects, \begin{equation} \left| k \right] ^a , \quad \left| p \right\rangle ^{ \dot{a} } \end{equation}
are two component spinors while $ \gamma_\mu$ is a 4x4 matrix. So its not even clear what the brakets mean. When we write the braket, \begin{equation} \left[ k | \gamma ^\mu | p \right\rangle \end{equation} what we really mean is that we pick out the Pauli matrix in the $ \gamma ^\mu $ with the correct index structure. For example, \begin{equation} \left[ k \right| ^a \left( \begin{array}{cc} 0 & \sigma ^\mu _{ a \dot{a} } \\ \bar{\sigma} ^\mu _{ \dot{a} a } & 0 \end{array} \right) \left| p \right\rangle^ {\dot{a}} = \left[ k \right| ^a \sigma ^{ \mu } _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \end{equation} with similar results for the other brakets.

Now getting back to your question. The $ \gamma ^\mu $ matrix is not invariant under complex conjugation: \begin{equation} \left( \gamma ^\mu \right) ^\dagger = \gamma _0 \gamma ^\mu \gamma _0 = \left( \begin{array}{cc} 0 & \bar{\sigma} ^\mu \\ \sigma ^\mu & 0 \end{array} \right) \end{equation} so all complex conjugation does is switch the positions of the $ \sigma ^\mu $ and $ \bar{\sigma} ^\mu $ matrices. Therefore, we can just omit the complex conjugation if we remember the ``pick Pauli matrix with the correct index structure'' prescription.

Then we have, \begin{equation} \left( \left[ k \right| ^a ( \gamma ^\mu ) _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \right) ^\ast = \left[ p \right| \gamma ^\mu \left| k \right\rangle \end{equation} where it is understood that now we are picking out the $ \bar{\sigma} ^\mu $ matrix instead of $ \sigma ^\mu $ in $ \gamma ^\mu $.

The problem is sloppy (but convenient) notation. The objects, \begin{equation} \left| k \right] ^a , \quad \left| p \right\rangle ^{ \dot{a} } \end{equation}
are two component spinors while $ gamma_\mu$ is a 4x4 matrix. So its even clear what the brakets mean. When we write the braket, \begin{equation} \left[ k | \gamma ^\mu | p \right\rangle \end{equation} what we really mean is that we pick out the $ \sigma ^\mu $ matrix in the $ \gamma ^\mu $ with the correct index structure. For example, \begin{equation} \left[ k \right| ^a \left( \begin{array}{cc} 0 & \sigma ^\mu _{ a \dot{a} } \\ \bar{\sigma} ^\mu _{ \dot{a} a } & 0 \end{array} \right) \left| p \right\rangle^ {\dot{a}} = \left[ k \right| ^a \sigma ^{ \mu } _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \end{equation} with similar results for the other brakets.

Now getting back to your question. The $ \gamma ^\mu $ matrix is not invariant under complex conjugation: \begin{equation} \left( \gamma ^\mu \right) ^\dagger = \gamma _0 \gamma ^\mu \gamma _0 = \left( \begin{array}{cc} 0 & \bar{\sigma} ^\mu \\ \sigma ^\mu & 0 \end{array} \right) \end{equation} so all complex conjugation does is switch the positions of the $ \sigma ^\mu $ and $ \bar{\sigma} ^\mu $ matrices. Therefore, we can just omit the complex conjugation if we remember the ``pick Pauli matrix with the correct index structure'' prescription.

Then we have, \begin{equation} \left( \left[ k \right| ^a ( \gamma ^\mu ) _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \right) ^\ast = \left[ p \right| \gamma ^\mu \left| k \right\rangle \end{equation} where it is understood that now we are picking out the $ \bar{\sigma} ^\mu $ matrix instead of $ \sigma ^\mu $ in $ \gamma ^\mu $.

The problem is sloppy (but convenient) notation. The objects, \begin{equation} \left| k \right] ^a , \quad \left| p \right\rangle ^{ \dot{a} } \end{equation}
are two component spinors while $ \gamma_\mu$ is a 4x4 matrix. So its even clear what the brakets mean. When we write the braket, \begin{equation} \left[ k | \gamma ^\mu | p \right\rangle \end{equation} what we really mean is that we pick out the $ \sigma ^\mu $ matrix in the $ \gamma ^\mu $ with the correct index structure. For example, \begin{equation} \left[ k \right| ^a \left( \begin{array}{cc} 0 & \sigma ^\mu _{ a \dot{a} } \\ \bar{\sigma} ^\mu _{ \dot{a} a } & 0 \end{array} \right) \left| p \right\rangle^ {\dot{a}} = \left[ k \right| ^a \sigma ^{ \mu } _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \end{equation} with similar results for the other brakets.

Now getting back to your question. The $ \gamma ^\mu $ matrix is not invariant under complex conjugation: \begin{equation} \left( \gamma ^\mu \right) ^\dagger = \gamma _0 \gamma ^\mu \gamma _0 = \left( \begin{array}{cc} 0 & \bar{\sigma} ^\mu \\ \sigma ^\mu & 0 \end{array} \right) \end{equation} so all complex conjugation does is switch the positions of the $ \sigma ^\mu $ and $ \bar{\sigma} ^\mu $ matrices. Therefore, we can just omit the complex conjugation if we remember the ``pick Pauli matrix with the correct index structure'' prescription.

Then we have, \begin{equation} \left( \left[ k \right| ^a ( \gamma ^\mu ) _{ a \dot{a} } \left| p \right\rangle ^{ \dot{a} } \right) ^\ast = \left[ p \right| \gamma ^\mu \left| k \right\rangle \end{equation} where it is understood that now we are picking out the $ \bar{\sigma} ^\mu $ matrix instead of $ \sigma ^\mu $ in $ \gamma ^\mu $.