When the base-$b$ number $11011_b$ is multiplied by $b-1$, then $1001_b$ is added, what is the result (written in base $b$)?
We can write out what $11011_b$ means in terms of powers of $b$: $$11011_b = b^4+b^3+b+1.$$Multiplying this by $b-1$ gives \begin{align*}
11011_b &= (b-1)b^4 + (b-1)b^3 + (b-1)b + (b-1) \\
&= b^5 - b^4 + b^4 - b^3 + b^2 - b + b - 1 \\
&= b^5 - b^3 + b^2 - 1.
\end{align*}Now $$1001_b = b^3 + 1,$$so when we add this to the result above, we get $b^5+b^2$, which is written in base $b$ as $\boxed{100100}$.

Instead of taking this algebraic approach, you can also think in terms of base-$b$ long arithmetic (note that each $(b-1)$ below represents a single digit): $$\begin{array}{r *5{c@{~}}c}
&& 1 & 1 & 0 & 1 & 1 \\
\times &&&&&& (b-1) \\
\hline
&& (b-1) & (b-1) & 0 & (b-1) & (b-1) \\
\\
\\
& \stackrel{1}{\phantom{(0)}} & \stackrel{1}{(b-1)} & (b-1) & \stackrel{1}{\phantom{(}0\phantom{)}} & \stackrel{1}{(b-1)} & (b-1) \\
+ &&& 1 & 0 & 0 & 1 \\
\hline
& 1 & 0 & 0 & 1 & 0 & 0
\end{array}$$Note that no carries are needed in the multiplication step, since $b-1$ is a digit in base $b$. Carries are needed in the addition step, since $(b-1)+1=10_b$.