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:
$\begin{align}
L &= a M\\
M &= a M + b P\\
P &= 1 + a M + b P
\end{align}$
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
$\begin{align}
P &= 1 + a a^*bP + b P\\
&= 1 + a^*bP
\end{align}$
... which has solution
$\begin{align}
P &= (a^*b)^*\\
&= 1 + (a + b)^*bP
\end{align}$
Now we can solve for $M$:
$\begin{align}
M &= a^*bP\\
&= a^*b(a^*b)^*\\
&= (a+b)^*b
\end{align}$
And for $L$:
$\begin{align}
L &= a M\\
&= a(a+b)^*b\\
\end{align}$
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?
$\begin{align}
a^{-1}M &= b + (a + b)^* b\\
&= (1 + (a + b)^*) b\\
&= (a + b)^* b\\
&= M
\end{align}$
Even if you miss this, you should find out by further beheading that $M$ and $N$ are indistinguishable.