This is very similar to the standard statement of the pumping lemma. Since you said you don't know how to begin, I suggest that you start by carefully review ing the proof of the standard pumping lemma that you will find in your textbook.
Then compare this statement to the standard pumping lemma to see exactly ehat the differences are, and see if you can make a small change to the standard proof to get the theorem that you need to prove here.
If you didn't understand the proof in your book, get a different book. The book by Michael Sipser is pretty good.
(Appending my comment.)
In the ordinary pumping lemma, the $uxv$ string gets the automaton into an accepting state. Here, it might not. It only gets the automaton into a state, say $S$, from which some additional input $z$ might get it to an accepting state. But if $uxv$ gets it to $S$, and $x$ goes around a loop, then $ux^iv$ will go around the loop $i$ times and get the automaton to $S$ also. And since $uxv$ gets the automaton to $S$, and $ux^iv$ gets it to $S$ also, then $ux^ivz$ must get it to the same state as $uxvz$. So if the automaton accepts $uxvz$, for some $z$, it must accept $ux^ivz$ also, and vice versa.