The above equation is from chapter 9.5 "Functional Quantization of the Spinor Field" of Peskin's and Schroeder's book $($page $305)$. I understand that the initial determinant equal to the final exponent as it is consistent with Eq.$7.98$, but how the intermediate step is got. I think it is related to Eq.$7.96$ but don't know how to explain it further.
This is more a sketch of the calculation than a detailed explanation, and there may be sign errors.
As mentioned at the end of page 304, one can alternatively use Feynman diagrams to evaluate $(9.76)$. To do so one uses two Grassmann-valued sources, say $\eta$ and $\bar{\eta}$, and use the generative functional: \begin{equation} Z[\eta,\bar{\eta}]=\int \mathcal{D}\bar{\psi} \mathcal{D}\psi\exp\left[ i\int d^4x \bar{\psi}(i\gamma \cdot D-m)+i\int d^4x (\bar{\eta}\psi+\bar{\psi} \eta)\right]. \tag{1} \end{equation} From here it is straightforward to evaluate this expression: \begin{equation} Z[\eta,\bar{\eta}] = e^{-e \int d^4x \frac{-i\delta}{\delta \bar{\eta}}\gamma^\mu A_\mu \frac{i\delta}{\delta \eta}}\det(i\gamma\cdot\partial-m)e^{-i\int d^4x \int d^4y\, \bar{\eta}(x) G(x-y) \eta(y)},\tag{2} \end{equation} where $G$ is the Green function of the free spinorial field. It is related to $S_F$, its propagator, by an $i$ factor. Expanding the first exponential in $(2)$ around $\eta=0$ and $\bar{\eta}=0$ we get: \begin{align} Z=&\det(i\gamma\cdot\partial-m)\left( 1-i\int d^4x \,\text{tr}\left[(-e \gamma^\mu A_\mu(x))S_F(0)\right] \right. \tag{3.a} \\ &-\frac{1}{2}\int d^4x d^4y\,\text{tr}\left[(-e \gamma^\mu A_\mu (x))(-e\gamma^\nu A_\nu(y))(iS_F(0))(iS_F(0))\right]\tag{3.b} \\ &\left.-2\times \frac{1}{2}\int d^4x d^4y\,\text{tr} \left[(-e \gamma^\mu A_\mu (x))(-e \gamma^\nu A_\nu(y))(iS_F(x-y))(iS_F(x-y))\right]+\cdots \right) \tag{3.c} \end{align} Normally, the traces appear when you do the calculation with explicit indices. Then, $(3.a)$ corresponds to $1$ plus the first tadpole in the intermediate step of $(9.79)$, $(3.b)$ to the product of the two tadpoles in this intermediate step, and finally $(3.c)$ corresponds to a loop attached with two sources (second loop in the intermediate step).
As I said, I have not checked if my signs are correct, so let me know if there are any errors. Nonetheless, it should be a good sketch of what is going on.
The RHS of eq. (9.79) is indeed just the Taylor expansion of the logarithm in eq. (9.78). Note that the factor $\frac{1}{n}$ from the Taylor expansion matches the (reciprocal) symmetry factor of the corresponding Feynman diagram on the RHS of eq. (9.79).
The middle expression of eq. (9.79) follows from the RHS via the linked-cluster theorem for connected Feynman diagrams.