As Geoff Robinson's answer makes clear, the Brauer map is injective when restricted to indecomposable $p$-permutation $kG$-modules with vertex $P$, because it agrees with the Green correspondence. This is stated in Theorem 3.4 of the paper by Broué mentioned in the question. Natalie's answer mentions Thévenaz's book, which has a careful proof of this fact.

If we allow non $p$-permutation modules (but still require fixed vertex $P$) then the map is not injective. For example, take $G = P = \langle g \rangle \cong C_8$ and let $k = \mathbf{F}_2$. Let $V_k$ be the unique $k$-dimensional indecomposable representation of $G$, for $1 \le k \le 8$. Then $V_3$ has a basis $v_1,v_2,v_3$ such that $v_i(g+1) = v_{i+1}$ for $i=1,2$ and $v_3(g+1) = 0$. Clearly $V_3^P = v_3$. Since

$$v_2g^2 = v_2 + v_2(1+g^2) = v_2 + v_2(1+g)^2 = v_2$$

and $\mathrm{Tr}_{\langle g^2 \rangle}^P v_2 = v_2(1+g) = v_3$, we have $\mathrm{Br}_P(V_3) = 0$. A similar argument shows that $\mathrm{Br}_P(V_5) = \mathrm{Br}_P(V_7) = 0$. But since $V_3$, $V_5$ and $V_7$ have odd dimension, they all have full vertex $P$.

There is a related example in Example 1.3.6 of the note of mine linked in the question.

As Geoff Robinson's answer makes clear, the Brauer map is injective when restricted to indecomposable $p$-permutation $kG$-modules with vertex $P$, because it agrees with the Green correspondence. This is stated in Theorem 3.4 of the paper by Broué cited in the question. Natalie's answer mentions Thévenaz's book, which also has a proof of this fact.

If we allow non $p$-permutation modules (but still require fixed vertex $P$) then the map is not injective. For example, take $G = P = \langle g \rangle \cong C_8$ and let $k = \mathbf{F}_2$. Let $V_k$ be the unique $k$-dimensional indecomposable representation of $G$, for $1 \le k \le 8$. Then $V_3$ has a basis $v_1,v_2,v_3$ such that $v_i(g+1) = v_{i+1}$ for $i=1,2$ and $v_3(g+1) = 0$. Clearly $V_3^P = v_3$. Since

$$v_2g^2 = v_2 + v_2(1+g^2) = v_2 + v_2(1+g)^2 = v_2$$

and $\mathrm{Tr}_{\langle g^2 \rangle}^P v_2 = v_2(1+g) = v_3$, we have $\mathrm{Br}_P(V_3) = 0$. A similar argument shows that $\mathrm{Br}_P(V_5) = \mathrm{Br}_P(V_7) = 0$. But since $V_3$, $V_5$ and $V_7$ have odd dimension, they all have full vertex $P$.

There is a related example in Example 1.3.6 of the note of mine linked in the question.