Building automaton from regular expression
You failed to notice that your $M$ and $N$ states are indistinguishable. Let's take a walk around the problem first. 



We can express this all in regular equations, in the notation of John Conway's Regular Algebra and Finite Machines: 

$\begin{align}
L &amp;= a M\\
M &amp;= a M + b P\\
P &amp;= 1 + a M + b P
\end{align}$

We can see that $L$


starts with an $a$;
remains in $M$ for as many $a$s as you like;
goes to $P$, the only terminal state, at the first sign of a $b$;
stays in the terminal $P$ for as many $b$s as you like;
returning to $M$ at the first sign of an $a$.


These equations translate immediately into a DFA. We can also solve them: 

$M = a M + b P$

... has solution

$M = a^*bP$

Substituting this into 

$P = 1 + a M + b P$

... we get

$\begin{align}
P &amp;= 1 + a a^*bP + b P\\
&amp;= 1 + a^*bP
\end{align}$

... which has solution

$\begin{align}
P &amp;= (a^*b)^*\\
&amp;= 1 + (a + b)^*bP
\end{align}$

Now we can solve for $M$:

$\begin{align}
M &amp;= a^*bP\\
&amp;= a^*b(a^*b)^*\\
&amp;= (a+b)^*b
\end{align}$

And for $L$:

$\begin{align}
L &amp;= a M\\
&amp;= a(a+b)^*b\\
\end{align}$



How do we solve the equations above? The solution to the equation

$X = \mathcal {x} + \mathcal y X$

... ,where $\mathcal {x}$ and $\mathcal {y}$ are regular expressions independent of $X$, is

$X = {\mathcal y}^* \mathcal {x} $



OK. What did you miss? 

$\begin{align}
a^{-1}M &amp;= b + (a + b)^* b\\
&amp;= (1 + (a + b)^*) b\\
&amp;= (a + b)^* b\\
&amp;= M
\end{align}$

Even if you miss this, you should find out by further beheading that $M$ and $N$ are indistinguishable.