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For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. By twisting the pair $(P_0,M_0)$ using $c$, we obtain a new pair of $k$-subgroups $(P_1,M_1)$ of $G$ that is conjugate to $(P_0,M_0)$ over an algebraic closure of $k$. It follows that $P_1$ is a minimal parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13 $P_1$ is conjugate to $P_0$ over $k$: $$P_1=g\cdot P_0\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $$M_1'=g\cdot M_0\cdot g^{-1},$$ then $M_1$ and $M_1'$ are two Levi subgroups of $P_1$, hence they are conjugate over $k$ in $P_1$ (a reference is needed). It follows that $M_1:=\,_c M_0$ is conjugate to $M_0$ over $k$ in $G$. This means that $\xi=1$, hence the kernel has only one element. A twisting argument shows that every fiber has only one element.