= Solution
The hypothesis says that $f$ has a zero of <order of a zero of a holomorphic function>[order] $k$ at $a$, so write
$$
f(z)=(z-a)^kg(z),
\qquad
g(a)\ne0,
$$
with $g$ holomorphic near $a$. Differentiating gives
$$
f'(z)=(z-a)^{k-1}\bigl(kg(z)+(z-a)g'(z)\bigr).
$$
Choose $\delta>0$ so small that the closed disc $\overline D(a,\delta)$ lies in the domain, $g$ never vanishes there, and the parenthesized factor never vanishes there. Thus $a$ is the only critical point of $f$ in that disc.
On the boundary circle, $f$ is nonzero. Set
$$
\varepsilon=\min_{|z-a|=\delta}|f(z)|>0.
$$
If $0<|b|<\varepsilon$, then $|b|<|f(z)|$ on the boundary. <Rouché's theorem> applied to $f$ and $f-b$ says that $f-b$ has the same number of zeros in $D(a,\delta)$ as $f$, namely $k$, counted with multiplicity.
None of these zeros is $a$ because $b\ne0$, and none is a critical point because $f'$ has no other zero in the disc. Every zero of $f-b$ is therefore simple. Hence there are exactly
$$
\boxed{k\text{ distinct }z\in D(a,\delta)\text{ with }f(z)=b}.
$$
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