= Solution
{c}
Put $q=p^n$. A finitely generated <module> over $A=k[X]/(X^q)$ is finite-dimensional over $k$, and multiplication by $X$ is a <nilpotent linear map> $T$ with $T^q=0$. Its <Jordan normal form> exists over $k$ because its minimal polynomial is a power of $X$. Each <Jordan block> has size $r\le q$ and is the <cyclic module> $M_r=k[X]/(X^r)$. Hence
$$
M\cong\bigoplus_{r=1}^{q}M_r^{\oplus m_r}.
$$
A <submodule> of $M_r$ is an ideal of $k[X]/(X^r)$, and its inverse image is an ideal of the <principal ideal domain> $k[X]$ containing $(X^r)$. The only possibilities are $(X^j)$ with $0\le j\le r$. Thus all <submodules> form the chain
$$
M_r\supset XM_r\supset\cdots\supset X^{r-1}M_r\supset0.
$$
Every successive quotient is the one-dimensional <simple module> on which $X$ acts as zero. This is its unique <composition series>, so $M_r$ is a <uniserial module> and is an <indecomposable representation>: two nonzero direct summands would be incomparable <submodules>. Only $M_1$ is simple.
For $G=\langle g\rangle$ of order $q$, the <group algebra> satisfies
$$
kG\cong k[T]/(T^q-1)\cong k[X]/(X^q),\qquad X=g-1,
$$
since $(T-1)^q=T^q-1$ in <characteristic> $p$. Consequently, up to isomorphism, $\boxed{M_1,\ldots,M_{p^n}}$ are precisely the finite-dimensional indecomposable modules, and $\boxed{M_1\text{ is the unique simple module}}$. The finite-dimensional convention follows from the finite-generation hypothesis; “exactly” counts isomorphism classes.
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