= Solution
Put $b=\operatorname{Br}_D(e)$. By equivariance of the <Brauer homomorphism>, $b$ is a <central idempotent> of $kK$, since $C_G(D)\le K\le N_G(D)$. The <Nagao module theorem> asserts that, if $eM=M$, then
$$
\boxed{M\downarrow_K=bM\oplus(1-b)M,}
$$
and every indecomposable summand of $(1-b)M$ is relatively projective for a $p$-subgroup of $K$ not containing $D$. Because $D$ is normal in $K$, its <module vertices> cannot contain $D$ even after $K$-conjugation.
Here is a direct proof. As an element of $(kG)^K$, $e-b$ is a linear combination of $K$-conjugation orbit sums supported outside $C_G(D)$. An orbit represented by $x$ has sum
$$
\operatorname{Tr}_{C_K(x)}^K(x).
$$
Its stabilizer $C_K(x)$ does not contain $D$. On $T=(1-b)M$, the element $e-b$ acts as the identity. Map the orbit-sum expression into $\operatorname{End}_k(M)$ by left multiplication, and compose on both sides with the $K$-linear projection onto $T$. It follows that
$$
\operatorname{id}_T\in\sum_{x\notin C_G(D)}
\operatorname{Tr}_{C_K(x)}^K\operatorname{End}_{kC_K(x)}(T).
$$
Replace each <centralizer> by a Sylow $p$-subgroup of it using trace transitivity and its invertible prime-to-$p$ index. Project this identity further onto an indecomposable summand $U$ of $T$. Its <endomorphism ring> is local by <Krull–Schmidt theorem>. A sum of nonunits cannot be the identity in a <local ring>, so one of the trace terms is a unit $u$ in $\operatorname{End}_{kK}(U)$. Multiplying the endomorphism inside that trace by $u^{-1}$ gives $\operatorname{id}_U$ as a single relative trace. The <D. Higman criterion> makes $U$ relatively projective for that <Sylow subgroup> $Q\le C_K(x)$, and $D\nleq Q$. This proves the assertion rather than just invoking the theorem's name.
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