= Solution
The <Jacobson radical> $J=J(R)$ is the intersection of all maximal right ideals, equivalently the largest ideal annihilating every simple right module. The <Artin–Wedderburn theorem> gives
$$
\boxed{R/J\simeq\prod_{a=1}^tM_{n_a}(k)}
$$
because $k$ is algebraically closed.
The descending chain $J\supseteq J^2\supseteq\cdots$ stabilizes since $R$ is finite-dimensional. If $J^m=J^{m+1}$, <Nakayama lemma> applied to the finite right module $J^m$ gives $J^m=0$. Thus $J$ is nilpotent.
For $f\in\operatorname{End}_R(P_i)$, the <Fitting lemma> gives
$$
P_i=\ker f^n\oplus\operatorname{im}f^n
$$
for large $n$. Indecomposability makes one summand zero, so $f$ is either invertible or nilpotent. In the latter case $1-f$ is invertible. This is the criterion that $\operatorname{End}_R(P_i)$ is a <local ring>.
Let $M=\bigoplus_iP_i$. Reduction modulo $MJ$ defines
$$
\theta:\operatorname{End}_R(M)\longrightarrow\operatorname{End}_R(M/MJ).
$$
Since $M/MJ=\bigoplus_iS_i$ and the $S_i$ are pairwise nonisomorphic simples,
$$
\operatorname{End}_R(M/MJ)\simeq\prod_i\operatorname{End}_R(S_i)\simeq k^n
$$
by <Schur lemma>. Arbitrary scalars on the direct summands lift to scalar identity maps on the $P_i$, so $\theta$ is surjective.
If $f\in\ker\theta$, then $f(M)\subseteq MJ$. A product of $r$ such maps sends $M$ into $MJ^r$, so $\ker\theta$ is nilpotent. A nilpotent ideal lies in the Jacobson radical, while the semisimplicity of the quotient $k^n$ gives the reverse inclusion. Hence
$$
\boxed{J(\operatorname{End}_R(M))=\ker\theta,\qquad
\operatorname{End}_R(M)/J\simeq k^n.}
$$
This is exactly the definition of a <basic algebra>.
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