= Solution
A finite-dimensional <group algebra> has <finite representation type of a group algebra> when there are only finitely many isomorphism classes of finite-dimensional <indecomposable modules>. The definition concerns all dimensions, not merely finitely many <modules> in each fixed dimension.
Let $e_B$ be the block <idempotent> and choose a maximal <Brauer pair> $(D,b_D)$: $b_D$ is a block <idempotent> of $kC_G(D)$ with $b_D\operatorname{Br}_D(e_B)\ne0$. The <inertial index of a block> is
$$
\boxed{e(B)=|N_G(D,b_D):DC_G(D)|,}
$$
where $N_G(D,b_D)$ is the stabilizer of the chosen pair. Changing the pair by conjugation does not alter this number. For cyclic $D$, it is a prime-to-$p$ divisor of $p-1$. Here $n\ge1$, as required for the order-$p$ <subgroup> $Q$ to exist.
Put $H=N_G(Q)$. Since $Q$ is characteristic in $D$, $N_G(D)\le H$, so <Green correspondence> applies. In fact the nonprojective <modules> in $B$ have a particularly simple correspondence. If $g\notin H$, a nontrivial intersection $D\cap{}^gD$ would contain the unique order-$p$ <subgroup> of both cyclic <groups>, giving $Q={}^gQ$ and hence $g\in H$, a contradiction. Thus every such intersection is trivial. The error <modules> in induction are projective. In restriction, an intersection $H\cap{}^gD$ cannot contain a conjugate of a nontrivial <subgroup> of $D$ by an element of $H$: such a conjugate contains $Q$, and that would again force ${}^gQ=Q$. The general <Green correspondence> therefore pairs all <indecomposable modules> with nontrivial <module vertex> contained in a conjugate of $D$. In particular it pairs the nonprojective <modules> of $B$ with the nonprojective <modules> of its Brauer correspondent $b$ in $H$; the block-compatible form of the correspondence selects $b$. Explicitly,
$$
M\downarrow_H=f(M)\oplus M_{\mathcal Y},\qquad
f(M)\uparrow^G=M\oplus P,
$$
where $P$ is projective and the restriction error family is $\mathcal Y=\{H\cap{}^gD:g\notin H\}$. One should not replace this restriction error term by projectives without an additional argument about $H$.
For the final structure calculation we use the general cyclic-defect structure theorem, which is permitted here as a general fact: the basic <algebra> of a split <block with cyclic defect group> is a <Brauer tree algebra>. Its tree has $e(B)$ edges, one simple <module> per edge, and exceptional multiplicity
$$
m=\frac{|D|-1}{e(B)}.
$$
For a one-edge tree the basic <algebra> presentation is $k[t]/(t^{m+1})$. This is the local presentation of the tree <algebra>, not an assertion that an arbitrary block with one simple <module> is automatically uniserial.
With $e(B)=1$ the tree has one edge and $m=p^n-1$. Thus the basic <algebra> of $B$ is
$$
A_0=k[t]/(t^{p^n}).
$$
It has exactly one simple <module>, $k=A_0/(t)$. Its <ideals> are precisely $(t^i)$: for any nonzero element, factor out its lowest power of $t$, after which the remaining factor is a unit. The regular projective therefore has the unique submodule chain
$$
A_0\supset tA_0\supset\cdots\supset t^{p^n-1}A_0\supset0,
$$
with each consecutive factor one-dimensional and simple. A <Morita equivalence> from $A_0$ to $B$ preserves simple <modules>, <projective covers>, submodule lattices and composition length. Hence
$$
\boxed{B\text{ has one simple module }S,\qquad P(S)\text{ is uniserial of length }p^n.}
$$
In the split setting one may also write $B\cong M_{\dim_k S}(A_0)$. The <projective cover> has $p^n$ <composition factors> all isomorphic to $S$, so its vector-space dimension is $p^n\dim_kS$, not generally $p^n$.
There is also an elementary reason why the uniserial conclusion follows once the one-simple-module conclusion and finite type are known. If the basic local <algebra> had $\dim J/J^2\ge2$, quotienting its radical square and all but two generators would give $k[x,y]/(x,y)^2$. The two-dimensional parameter family from Question 3(b) would then give infinitely many indecomposables. Thus $J/J^2$ is one-dimensional. <Nakayama lemma> makes $J$ generated by one element $t$, so its powers yield a chain and the basic <algebra> is $k[t]/(t^r)$. The cyclic-defect calculation above identifies $r=p^n$. This explains why finite representation type rules out branching in this <projective cover>.
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