The question:
Let it be $L$ a regular language.
few definitions:
$p(L)$-the minimum natural number so that $L$ fulfills the pumping lemma.
$n(L)$- minimal NFA that accepts $L$.
$m(L)$- $Rank(L)$, the number of equivalence classes in $L$.
Let it be integer $k>0$. find an example for a language $L$ so that: $p(L)=n(L)=m(L)=k$.
My attempt:
At start, thought of $L=\{w: |w|\bmod k=0\}$, $\Sigma=\{ a\}$.
But then I realized that $p<k$.
And also- for every $L$ that I choose here, I don't know how exactly to prove that the $NFA$ which accepts it is minimal. I know how to do this only on $NFA$'s with one or two states, but not on $NFA$'s with unknown number of states.
How can I solve this?
Your answer is correct.
If $p<k$ then for $w=a..a$ such that $|w|=p<k$, $w\in L$. But as you defined $L$ it is not possible.
In your case $p=k$ because the finite automaton that accepts $L$ has exactly $k$ states. So every word of size $k$ ensures that you pass through at least one state twice. Thus the pumping lemma holds for $p=k$.
Notice that for a word of size $k$ you will pass through $k+1$ states as the start state is "given" for free.