= Solution
Put $N=N_G(D)$. For an algebra on which a group acts by conjugation, write
$$
A_D^H=\operatorname{Tr}_D^H(A^D),
\qquad
\operatorname{Tr}_D^H(a)=\sum_{h\in H/D}hah^{-1},
$$
for its <transfer ideal of conjugation-fixed elements>. In the diagram, $\tau=\operatorname{Tr}_D^G$, $\tau'=\operatorname{Tr}_D^N$, the upper map is $\beta=\operatorname{Br}_D$, and $\beta'$ is the restriction of $\operatorname{Br}_D$ from $(kG)^G$ to $(kG)_D^G$. Since $N$ normalizes both $D$ and $C_G(D)$, the lower map lands in $(kC_G(D))_D^N$.
For $a\in(kG)^D$, let $D$ act on the left cosets $G/D$. A coset $gD$ is fixed precisely when $g^{-1}Dg\leq D$, hence, because the two groups have the same order, precisely when $g\in N$. Every nonfixed orbit has size divisible by $p$. Moreover, after applying $\operatorname{Br}_D$, all summands indexed by one $D$-orbit are equal: conjugation by an element of $D$ acts trivially on $kC_G(D)$. Those orbits therefore contribute zero in characteristic $p$, while the fixed cosets contribute the trace over $N/D$. Consequently
$$
\beta'\tau(a)
=\operatorname{Br}_D\!\left(\operatorname{Tr}_D^G(a)\right)
=\operatorname{Tr}_D^N\!\left(\operatorname{Br}_D(a)\right)
=\tau'\beta(a).
$$
This is the <Brauer morphism and relative trace> identity, so \b[the diagram commutes].
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