Apply Ito's lemma to $\ln M_t$, we obtain that
\begin{align*}
d\ln M_t &= \frac{1}{M_t} dM_t -\frac{1}{2} \frac{1}{M_t^2} d\langle M, M\rangle_t\\
&=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \frac{1}{M_t^2}\left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt\\
&=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right) dt.\tag{Eq. 1}
\end{align*}
Here, since $dM_t=M_t\big[-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t\big]$,
\begin{align*}
d\langle M, M\rangle_t = \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt.
\end{align*}
From $(\textrm{Eq.} 1)$,
\begin{align*}
\ln M_t - \ln M_0 &= \int_0^t\left[-\frac{\mu}{\sigma} dW_s + \gamma_s dB_s -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_s^2\right) ds\right]\\
&=-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds.
\end{align*}
That is,
\begin{align*}
M_t = M_0 \exp\left(-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds\right).
\end{align*}