= Solution
{c}
Use the source's <charge-conjugation matrix> convention $C\gamma^\mu C^{-1}=-(\gamma^\mu)^T$. It implies
$$
C^{-1}(\gamma^\mu)^TC=-\gamma^\mu,\qquad
C^{-1}K(p)^TC=K(-p).
$$
A <trace> equals the <trace> of its <matrix transpose>, which reverses the order of all factors. Inserting the displayed identities gives
$$
\begin{aligned}
T_A(\ell)&=\operatorname{tr}[\gamma^\mu K(\ell+k_1)\gamma^\nu K(\ell-k_3)\gamma^\rho K(\ell)]\\
&=-\operatorname{tr}[K(-\ell)\gamma^\rho K(-\ell+k_3)\gamma^\nu K(-\ell-k_1)\gamma^\mu]\\
&=-\operatorname{tr}[\gamma^\mu K(-\ell)\gamma^\rho K(-\ell+k_3)\gamma^\nu K(-\ell-k_1)].
\end{aligned}
$$
There is one minus sign from each of the three photon <interaction vertices>, and none from the transformed <Dirac propagators>. In the last line, change <loop momentum> to $\ell'=-\ell-k_1$. Since $k_1+k_3=-k_2$, its three <Dirac propagators> become $K(\ell'+k_1)$, $K(\ell'-k_2)$, $K(\ell')$, exactly those of the reversed orientation. Hence
$$
\boxed{\mathcal M_A^{\mu\nu\rho}=-\mathcal M_B^{\mu\nu\rho},\qquad
\mathcal M_{\rm one\ loop}^{\mu\nu\rho}=0.}
$$
This is <Furry's theorem>. The <regularization in quantum field theory> matters: the individual <Feynman integrals> are superficially divergent, so an unregulated shift of a divergent integral is not a proof. A common <dimensional regularization> preserves <charge conjugation> and the vector <Ward identity>, making the change of variables legitimate. No <chirality matrix> occurs, so the axial triangle <anomaly (physics)> is not involved.
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