A state doesn't recognise a language. A DFA recognises a language by accepting exactly the set of words in the languages and no others. A DFA has many states.
If there is a regular language L, which can be modelled by the Pumping Lemma, it will have a property n.
For a DFA with s states, in order for it to accept L, s must be >= n.
The last line merely states that there are some languages in which s is greater than n, rather than equal.
This is probably more suited for https://cstheory.stackexchange.com/ or https://cs.stackexchange.com/ (not quite sure of the value of both myself).
Note: Trivially, not all DFA's with sufficient states will accept the language. Also, the fact that a language passes the pumping lemma doesn't mean it's regular (but failing it means definitely isn't).
Note also, the language changes between FA and DFA - this is a bit lax, but because NDFAs have the same power as DFAs and DFAs are easier to write and understand, DFAs are used for the proof.
Edit I'll give an example of a regular language, so you can see an idea of u,v,w,z and n. Then we'll discuss s.
L = xy^nz, n > 2 (i.e. xyyz, xyyyz, xyyyyz)
u = xy
v = y
w = z
n = 4
Hence:
|z| = 3: xyz (i = 0) Not in L, but |z| < n
|z| = 4: xyyz (i = 1)
|z| = 5: xyyyz (i = 2)
etc
Hence, it's modelled by the Pumping Lemma. A DFA would need at least 4 states. So let's think of one.
-> State 1: x
State 1:
-> State 2: y
State 2:
-> State 3: y
State 3:
-> State 3: y
-> State 4: z
State 4:
Accepting state
Terminating state
4 states, as expected. Not possible to do it in less, as expected by n = 4. In this case, the example is quite simple so I don't think you can build one with more than 4 states (but that would be okay if it were needed).
