= Solution
Let $B=kGe$ have defect <group> $D$. Use two vertex-and-source facts, with their relevant content made explicit. Every <indecomposable module> in $B$ has a <module vertex> conjugate into $D$. Also $B$, as a $k[G\times G]$-module with $(g,h)a=gah^{-1}$, is an indecomposable permutation-module summand with <module vertex> $\Delta D=\{(d,d):d\in D\}$ and trivial <module source>. The second fact is the bimodule characterization of a <defect group of a block>; the first follows by applying its relative-$\Delta D$ induction splitting to $B\otimes_{kG}M=M$.
If $D$ is cyclic, every <module source> for every <module vertex> $Q\le D$ occurs among the finitely many <indecomposable modules for a cyclic p-group>. Each <module> in the block is a summand of an induction of one of these finitely many <module sources>. There are finitely many <subgroups> $Q$ and each <induced representation> has finitely many indecomposable summands by <Krull–Schmidt theorem>. Thus \b[cyclic defect implies finite representation type of the block].
For the converse, the restriction $B\downarrow_{D\times D}$ has the regular $kD$-bimodule as a <direct summand>. To see this from <module vertices> and <module sources>, its <Brauer quotient of a module> at $\Delta D$ is nonzero. Its summands are <permutation modules> for the $p$-group $D\times D$ and have <module vertices> of order at most $|D|$. A summand detected at $\Delta D$ therefore has <module vertex> $\Delta D$ and trivial <module source>, hence is
$$
\operatorname{Ind}_{\Delta D}^{D\times D}k\cong kD.
$$
Here a transitive <permutation module> for a $p$-group is indecomposable: it has one-dimensional <head of a module> over the local <group algebra> and cannot split into two nonzero <modules>. This also explains the Brauer-quotient detection used in the argument.
For any $kD$-module $U$, tensor that split bimodule inclusion with $U$. It gives
$$
U\mid\operatorname{Res}_D^G(B\otimes_{kD}U),
$$
where $\mid$ denotes being a <direct summand>. If $B$ has only finitely many <indecomposable modules> $M_1,\ldots,M_s$, each $B\otimes_{kD}U$ is a sum of these. Therefore every indecomposable $U$ must occur among the summands of the finite list $M_i\downarrow_D$. Thus $kD$ has finite representation type.
A noncyclic $p$-group has a quotient $C_p\times C_p$: its <Frattini quotient> has rank at least two, since a one-generator <Frattini quotient> forces the <group> itself to be cyclic. Inflation along this quotient embeds all the <indecomposable modules> of that elementary abelian <group>, preserving isomorphisms and indecomposability. Part (b) rules out finite type when $k$ is infinite.
The conclusion remains valid if the splitting <field> $k$ is finite. For completeness, replace the scalar parameter in part (b) by an $m\times m$ companion matrix $T_f$ of an irreducible polynomial $f$. On $k^m\oplus k^m$ use
$$
x=\begin{pmatrix}0&0\\I&0\end{pmatrix},\qquad
y=\begin{pmatrix}0&0\\T_f&0\end{pmatrix}.
$$
An endomorphism commuting with these has form $\left(\begin{smallmatrix}A&0\\C&A\end{smallmatrix}\right)$ with $AT_f=T_fA$. Its <endomorphism ring> has a square-zero <ideal> of the $C$-matrices and quotient $k[T_f]\cong k[t]/(f)$, a <field>, so it is local and the <module> is indecomposable. Finite <fields> have irreducible polynomials of arbitrarily large degrees, giving infinitely many dimensions. This also excludes noncyclic $D$ in that case. Altogether,
$$
\boxed{B\text{ has finite representation type}\quad\Longleftrightarrow\quad D\text{ is cyclic}.}
$$
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