= Solution
{c}
There is a missing hypothesis in the original PDF: $k$ must be a <splitting field for finite group representations>. For example,
$$
\mathbb F_2C_3\cong\mathbb F_2[X]/(X^3-1)\cong\mathbb F_2\times\mathbb F_4,
$$
because $X^3-1=(X-1)(X^2+X+1)$ and the quadratic factor is irreducible. This algebra has two <simple modules>, whereas $C_3$ has three conjugacy classes, all $2$-regular. Thus the printed assertion for an arbitrary <field> is false.
Under the intended splitting hypothesis, fix compatible lifts defining <Brauer characters>. Let $\mathcal K_{p'}$ be the set of conjugacy classes of <p-regular elements>. Character additivity and the tensor-product formula define the unital algebra homomorphism
$$
\Phi:\mathbb C\otimes_{\mathbb Z}R_k(G)\longrightarrow\prod_{C\in\mathcal K_{p'}}\mathbb C,\qquad z\otimes[V]\longmapsto\bigl(z\chi_V(g_C)\bigr)_C.
$$
We use the <Brauer character basis theorem>: over a splitting <field> the irreducible <Brauer characters> form a complex basis of the <class functions> on the p-regular conjugacy classes. Therefore $\Phi$ maps the basis $1\otimes[S]$ bijectively to a basis and is an algebra isomorphism. Comparing dimensions gives
$$
\boxed{\mathbb C\otimes R_k(G)\cong\mathbb C^{\mathcal K_{p'}},\qquad\#\{\text{simple }kG\text{-modules}\}=|\mathcal K_{p'}|\quad(k\text{ splitting})}.
$$
Isomorphism classes are understood in the count.
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