Transposition of spinors Suppose we have two 4-components Dirac spinors, that is two non commuting objects, $\psi$ and $\chi$. We know that:
$ \bar{\psi} \chi= - \chi \bar{\psi} $
$ \bar{\psi} = \psi^{\dagger} \gamma_0 $
$\dagger=T*$
The question is the following. What is $( \bar{\psi} \chi)^{T } $ ? For $T$ I mean taking the transpose, $\dagger$ is the adjoint, $*$ the complex conjugate.
Explicitly, is $ (\bar{\psi} \chi )^{T }=\chi^{T }\bar{\psi} ^{T } $ or $ (\bar{\psi} \chi )^{T }=- \chi^{T }\bar{\psi} ^{T } $ ?
And what can we say about $( \bar{\psi} \chi)^{\dagger} $? Is $ (\bar{\psi} \chi )^{\dagger}=\chi^{\dagger}\bar{\psi} ^{\dagger} $ or $ (\bar{\psi} \chi )^{\dagger}=- \chi^{\dagger}\bar{\psi} ^{\dagger} $ ?
Please, make the explanation clear, not only giving the correct answer.
 A: There're a few things that should be clarified. First of all, are you talking about classical spinors?
If they are classical spinors, then they take values in Grassmann-odd numbers, and so its transpose should be
\begin{align}
(\bar{\psi}\chi)^{T}&=(\psi^{\dagger}\gamma^{0}\chi)^{T} \\
&=\left[(\psi^{\ast})^{T}\gamma^{0}\chi\right]^{T} \\
&=-\bar{\psi}\chi,
\end{align}
where in the last step, one has
\begin{align}
\bar{\psi}\chi=(\psi^{\ast})^{T}\gamma^{0}\chi &=\sum_{i=0}^{3}\sum_{j=0}^{3}(\psi^{\ast})_{i}(\gamma^{0})_{ij}(\chi)_{j} \\
&=-\sum_{i=0}^{3}\sum_{j=0}^{3}(\chi)_{j}(\psi^{\ast})_{i}(\gamma^{0})_{ij} \\
&=-\sum_{i=0}^{3}\sum_{j=0}^{3}(\chi)_{i}(\psi^{\ast})_{j}(\gamma^{0})_{ji} \\
&=-\sum_{i=0}^{3}\sum_{j=0}^{3}(\chi)_{i}\left[(\gamma^{0})^{T}\right]_{ij}(\psi^{\ast})_{j} \\
&=-\chi^{T}(\gamma^{0})^{T}\psi^{\ast}=-\chi^{T}\bar{\psi}^{T}\equiv-(\bar{\psi}\chi)^{T}.
\end{align}

Now, the problem has to do with the involution of the Grassmann algebra $\Lambda_{\infty}$. To be specific, suppose you have a bunch of anti-commuting elements $\theta_{i}$, i.e.
$$\theta_{i}\theta_{j}+\theta_{j}\theta_{i}=0.$$
Then, each element (also known as the supernumber) in $\Lambda_{\infty}$ takes the form
$$z=z_{B}+\sum_{k=1}^{\infty}\frac{1}{k!}C_{i_{1}\dots i_{k}}\,\theta_{i_{1}}\cdots\theta_{i_{k}},$$
where $z_{B}\in\mathbb{C}$ is its body part, and $C_{i_{1}\dots i_{k}}\in\mathbb{C}$ is the coefficient of its soul part.
The involution (complex conjugation) $\ast$ in $\Lambda_{\infty}$ is defined as an action satisfying the following rules: 
$1.\quad(\theta_{i})^{\ast}=\theta_{i}$. 
$2.\quad(\alpha z)^{\ast}=\bar{\alpha}z^{\ast}$, where $\alpha\in\mathbb{C}$. 
$3.\quad(z+w)^{\ast}=z^{\ast}+w^{\ast}$. 
$4.\quad(zw)^{\ast}=w^{\ast}z^{\ast}$. 
$5.\quad(z^{\ast})^{\ast}=z$.

Back to your question, the classical Dirac spinors should take values in the infinite dimensional $\Lambda_{\infty}$. So you would expect
\begin{align}
(\bar{\psi}\chi)^{\dagger}&=\left[(\bar{\psi}\chi)^{T}\right]^{\ast}=\left[\chi^{T}\bar{\psi}^{T}\right]^{\ast} \\
&=\left(\sum_{i=0}^{3}\sum_{j=0}^{3}(\chi)_{i}\left[(\gamma^{0})^{T}\right]_{ij}(\psi^{\ast})_{j}\right)^{\ast} \\
&=\sum_{i=0}^{3}\sum_{j=0}^{3}\left[(\chi)_{i}\left[(\gamma^{0})^{T}\right]_{ij}(\psi^{\ast})_{j}\right]^{\ast} \\
&=\sum_{i=0}^{3}\sum_{j=0}^{3}(\psi)_{j}\left[(\gamma^{0})^{\dagger}\right]_{ij}(\chi^{\ast})_{i} \\
&=\sum_{i=0}^{3}\sum_{j=0}^{3}(\psi)_{j}(\gamma^{0})_{ij}(\chi^{\ast})_{i} \\
&=-\sum_{i=0}^{3}\sum_{j=0}^{3}(\chi^{\ast})_{i}(\gamma^{0})_{ij}(\psi)_{j}\equiv-\chi^{\dagger}\gamma^{0}\psi=-\bar{\chi}\psi.
\end{align}
Notice that in the last step one also has $\bar{\psi}^{\dagger}=(\psi^{\dagger}\gamma^{0})^{\dagger}=\gamma^{0}\psi$, and so you should obtain
$$(\bar{\psi}\chi)^{\dagger}=-\chi^{\dagger}\bar{\psi}^{\dagger}.$$
A: EDIT:
The correct answer is at: https://physics.stackexchange.com/a/458455/50207
Leaving the (wrong) answer below as it can be instructive.

I've clarified my doubt. The expression
$ (\bar{\psi} \chi )^{T }=\chi^{T }\bar{\psi} ^{T } $
is simply the definition of trasposition (the same argument for the adjoint), independently on the nature of the objects in the parenthesis. What is subtle is that, the following is true:
$ \bar{\psi} \chi =-\chi^{T }\bar{\psi} ^{T } =-(\bar{\psi} \chi )^{T }$
The minus in the first step arise only due to the anticommuting nature of the variables.
