Apply part (b) with $P=P_i$ and $U=S_j$. A homomorphism $P_i\to S_j$ kills $J(P_i)$ and therefore factors through the head $P_i/J(P_i)\cong S_i$. By <Schur lemma> over the splitting field,
$$
\dim\operatorname{Hom}_{kG}(P_i,S_j)=\delta_{ij}.
$$
Consequently the two stated bases satisfy the <Duality of simple and projective Brauer characters>:
$$
\boxed{\langle\chi_{P_i},\chi_{S_j}\rangle=\delta_{ij}.}
$$

The form is nondegenerate directly: if $\phi\ne0$, then
$$
\langle\phi,\phi\rangle
=\frac1{|G|}\sum_{g\ p\text{-regular}}|\phi(g)|^2>0.
$$
Equivalently, in coordinates indexed by the p-regular conjugacy classes it is a positive diagonal <Hermitian form> with weights $1/|C_G(x)|$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 2 / c / Solution

{c}

If $S_i$ are the simple modules of a split group algebra and $P_i$ are their <projective covers>, then
$$
\langle\chi_{P_i},\chi_{S_j}\rangle_{p'}=\delta_{ij}.
$$

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 Incoming links: Duality of simple and projective Brauer characters