If the eigenvalues of $A,M$ are all distinct $1,\lambda_2,\lambda_3$ and $1,m\lambda_2,m\lambda_3$, $B$ is any $3\times 3$ matrix with positive entries(0 is allowed), If the rank of $K=[B\hspace{0.5cm} AB\hspace{0.5cm} A^2B]$ is $3$ can we say anything about the rank of
$L=[B\hspace{0.5cm} MB\hspace{0.5cm} M^2B]$
No, you cannot say anything. Take e.g. $\lambda_2=2$ and $\lambda_3=3$, and the eigenvectors of $A$ to be the unit vectory. For $B=(1,1,1)'$ you have rank of $K$ equal 3.
If you now set $M$ to have the same eigenvalues ($m=1$), but with respective eigenvectors $(1,1,1)', (1,0,0)', (0,1,0)'$ you have rank of $L$ = 1.
Of course $L$ can have rank 3 if you set $M=A$, the construction of an example of rank $2$ is left to the reader.
So even with $m=1$ you cannot say anything.
(I used $B$ as a vector as mentionned in the comments. It's the same for matrices by using three copies of $B$).