Minimum pumping length of concatenation of two languages
For a language $L$, denote $P(L)$ as the minimum pumping length of $L$. It is implied that $P(L)\ge1$.

Proposition. Let $A,B$ be two languages with finite minimum pumping lengths. We have $P(AB)\le P(A)+P(B)-1$. Moreover, there exist regular $A,B$ such that the equality holds.

Proof. Let $w=w_Aw_b\in AB$, where $w_A\in A, w_B\in B$. Assume $|w|\ge P(A)+P(B)-1$. Then either $|w_A|\ge P(A)$ or $|w_B|\ge P(B)$; otherwise, $$|w|=|w_A|+|w_B|\le P(A)-1 + P(B)-1\lt P(A)+P(B)-1,$$
which contradicts the assumption. 

Now there are two cases.


$|w_A|\ge P(A)$. Since $P(A)$ is a pumping length for $A$, $w_a\in A$ can be written as $w_a=xyz$, satisfying the following conditions:


$|y|\geq 1$ 
$|xy|\leq P(A)$
$\forall n\geq 0, xy^{n}z\in A$


Hence, $w$ can be written as $w=xyz_B$, where $z_B=zw_B$, satisfying the following conditions:  


$|y|\geq 1$ 
$|xy|\leq P(A) + P(B)-1$ (since $P(B)\ge1$)
$\forall n\geq 0, xy^{n}z_B=(xy^nz)w_B\in AB$.


That is, $w$ can be pumped with pumping length $P(A)+P(B)-1$.
$|w_B|\ge P(B)$. This case is similar to the case above.


Combining the two cases, we have proved $P(A)+P(B)-1$ is a pumping length for $AB$. 

Let $A=\{\epsilon,\alpha, \alpha^2, \cdots, \alpha^{n-1}, \beta^n, \beta^{2n}, \beta^{3n}, \cdots, \}$, which is a regular language of minimal pumping length $n$.  Let $B=\{\epsilon,\alpha, \alpha^2, \cdots, \alpha^{m-1}, \beta^m, \beta^{2m}, \beta^{3m}, \cdots\}$, which is a regular language of minimal pumping length $m$. Since $\alpha^{n-1}\alpha^{m-1}$, which is a word in $AB$ of length $n+m-2$, cannot be pumped, $AB$ is a regular language of minimal pumping length $n+m-1$.