NFA accepting $L=\{wbbav \;|\; w \in \{a,b\}^*, v \in \{a,b\}^+, v\; has\; suffix\; a\}$
I will address your second question, namely, whether there is an NFA with fewer states. The appropriate technique here is fooling set:


  Let $L$ be a language, and let $x_1,\ldots,x_n,y_1,\ldots,y_n$ be words such that $x_iy_i \in L$ but for every $i \neq j$, either $x_iy_j \notin L$ or $x_jy_i \notin L$. Then every NFA for $L$ contains at least $n$ states.


(The proof is a simple exercise.)

In our case, we take
$$
\begin{array}{c|cc}
i &amp; x_i &amp; y_i \\\hline
1 &amp; bbaa &amp; \epsilon \\
2 &amp; bba &amp; a \\
3 &amp; bb &amp; aa \\
4 &amp; b &amp; aaa \\
5 &amp; \epsilon &amp; bbaa
\end{array}
$$
For all $i$ we have $x_iy_i = bbaa \in L$. Conversely, if $i &lt; j$ then $|x_jy_i| &lt; 4$ and so $x_jy_i \notin L$.

In fact, we have proven the following more general result:


  If the minimal length of a word in $L$ is $\ell$, then every NFA for $L$ contains at least $\ell+1$ states.