NPDA for $\{w : w \in \{a,b\}^*,n_a(w)\geq n_b(w)+1 \}$
"Is there a two-state NPDA that accepts the same language?"

Yes, every PDA can be replaced by one that has two states, one accepting and one non-accepting. In general one uses three steps to prove this formally:


Every PDA (with empty stack acceptance) can be transformed into a CFG. The construction is sometimes known as the triplet construction: a single nonterminal of the CFG codes two states and a stack symbol of the PDA. 
Every CFG can be transformed into a PDA with a single state and empty stack acceptance. That equivalence is quite straghtforward.
PDA with empty stack acceptance are equivalent to PDA with final state acceptance. But, where the stack-PDA may have a single state, the state-PDA usually uses one more, for acceptance.


Always, when discussing PDA, be explicit on the acceptance mode. (But that might be a silly request when your textbook considers only one of these types.)

PS. By the way, your machine seems to have the basic counting ideas implemented.