The expression $\sum_{k=1}^{\infty}p(1-p)^{k-1}\cdot e^{-\frac{k^c}{a}}$ is the average of $e^{-\frac{k^c}{a}}$ over a geometric distribution, such that $0\leq p \leq 1$, $c \geq 0$, and $a\geq 0$ (i.e. it converges). Is there a closed-form expression for this sum?
There are extreme cases where there is such an expression. There is the trivial case for $p = 0$. Setting $c = 0$ yields $e^{-\frac{1}{a}}$ and $c = 1$ yields $\frac{p}{e^{\frac{1}{a}}-1+p}$.
Numerical work for specific instances of $p$ and $c$ (such as $p = \frac{1}{2}$ and $c = 2$ did not return anything usable). Dropping the geometric distribution itself gives back for c = 1 and c = 2 that $\sum_{k=1}^{\infty}e^{-\frac{k}{a}} = \frac{1}{e^{\frac{1}{a}}-1}$ and $\sum_{k=1}^{\infty}e^{-\frac{k^2}{a}} = \frac{1}{2}\left(\theta\left(3, 0, e^{-\frac{1}{a}}\right)-1\right)$, where $\theta$ is the elliptic theta function.
Having a closed-form expression for integer $c$ (or just $c = 2$) would already be very nice, however.
Case $k=2$ is a so-called "partial theta function".
Consider this Jacobi theta function: $$ \vartheta_3(z,q) := \sum_{k=-\infty}^{\infty} q^{k^2} e^{i 2 k z} \;. $$ Then $$ p\,\vartheta_3\left(\frac{\log(1-p)}{2i}\, , \, e^{-1/a}\right) = \sum_{k=-\infty}^\infty p(1-p)^k e^{-k^2/a}\;. $$ The corresponding "partial theta function" is then what you want: $$ p\,\tilde\vartheta_3\left(\frac{\log(1-p)}{2i}\, , \, e^{-1/a}\right) = \sum_{k=1}^\infty p(1-p)^k e^{-k^2/a}\;. $$ Unfortunately, partial theta functions do not have the nice properties of theta functions, and are almost never seen in the literature.