= Solution
{c}
Completeness permits <idempotent lifting> from $kG$ to $\mathcal OG$. In particular, each <projective cover> $P_j$ lifts to a finite projective $\mathcal OG$-lattice $\widetilde P_j$; one can lift an idempotent presenting it as a summand of a <finite free module>. Set $Q_j=K\otimes_{\mathcal O}\widetilde P_j$.
For a simple $S_\ell$, every map $P_j\to S_\ell$ factors through its simple head $S_j$. Splitting and <Schur lemma> give $\dim_k\operatorname{Hom}_{kG}(P_j,S_\ell)=\delta_{j\ell}$. Exactness of this Hom functor along a <composition series> therefore gives
$$
\dim_k\operatorname{Hom}_{kG}(P_j,\overline W_i)=d_{ij}.
$$
Part (iii) identifies this with $\dim_K\operatorname{Hom}_{KG}(Q_j,V_i)$. By <Maschke's theorem> and splitting, $KG$ is split semisimple, so
$$
Q_j\cong\bigoplus_i V_i^{\oplus d_{ij}}.
$$
The reductions of any two integral forms of the same ordinary module have identical <Brauer characters>, hence identical composition multiplicities. We may thus reduce a direct-sum form for the displayed decomposition instead of $\widetilde P_j$. In the <modular representation ring> this gives
$$
[P_j]=\sum_i d_{ij}[\overline W_i]=\sum_\ell\left(\sum_i d_{i\ell}d_{ij}\right)[S_\ell].
$$
Comparing the simple basis coefficients proves $c_{\ell j}=\sum_i d_{i\ell}d_{ij}$, or $\boxed{C=D^TD}$.
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