You can proceed to some extent, although not fully, by using the Leibniz formula
$$
(f g)^{(n)}|_{t=0}=\sum_{p=0}^{n}\left(\begin{array}{l}{n} \\ {p}\end{array}\right) f^{(n-p)}|_{t=0} ~g^{(p)}|_{t=0}
$$
Your $f(t)=e^{kt}$ and therefore $f^{(n-p)}|_{t=0}=k^{n-p}$. I couldn't find a simple form for the $p$th derivative of your $g(t)=
\left(\frac{\sinh (t / n)}{\cosh (t / n)-a }\right)
$, other than the fact that an even number of derivatives vanish at $t=0$. Therefore,
$$
\frac{d^n}{dt^n} \left(e^{k t} \times\frac{\sinh (t / n)}{\cosh (t / n)-a )}\right)\bigg|_{t=0}=\sum_{p=1~, ~p\in 2\mathbb{Z}^++1}^n \binom{n}{p}k^{n-p} \frac{d^p}{dt^p} \left(\frac{\sinh (t / n)}{\cosh (t / n)-a }\right)\bigg|_{t=0}~.
$$
[Not sure whether this is a question for Mathematica though. At times you can get lucky – given a sequence of functions as a list, you can get the $n$th one in the sequence using FindSequenceFunction
but this specific case is too complicated.]
Edit
We can go a step further with Leibniz rule :
$$\frac{d^p}{dt^p} \left(\frac{\sinh (t / n)}{\cosh (t / n)-a }\right)\bigg|_{t=0}=\sum_{q=0~, ~q\in 2\mathbb{Z}^++1}^{q=p}\binom{p}{q} n^{q-p} \frac{d^q}{dt^q} \left(\frac{1}{\cosh (t / n)-a }\right)\bigg|_{t=0}$$
Therefore,
$$
\frac{d^n}{dt^n} \left(e^{k t} \times\frac{\sinh (t / n)}{\cosh (t / n)-a )}\right)\bigg|_{t=0}=\sum_{p=1~, ~p\in 2\mathbb{Z}^++1}^n \binom{n}{p}k^{n-p} \sum_{q=0~, ~q\in 2\mathbb{Z}^++1}^{q=p}\binom{p}{q} n^{q-p} \frac{d^q}{dt^q} \left(\frac{1}{\cosh (t / n)-a }\right)\bigg|_{t=0}.
$$
This is probably as general as it gets for arbitrary $n$.
D
can take arbitrary $ n $ times derivative, where $ n $ need not be a specific number, but sometimes the result is not very much useful. $\endgroup$