jmc

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 $$\begin{array}{rl}11011_b& =(b-1)b^4+(b-1)b^3+(b-1)b+(b-1)\\ & =b^5-b^4+b^4-b^3+b^2-b+b-1\\ & =b^5-b^3+b^2-1.\end{array}$$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 $(b-1)+1=10_b$.