Assume that $AB<AC$ Call the circle with radius $M$ and radius $MD$; $W$. let $W\cap BC=E, D$ obviously $E$ is tangency point of A_excircle with $BC$. Let $S, T$ feet of $B, C$ on $AI$ note that $ ISDB, IBI_AC$ and $ CTI_A$ are cyclic so $\angle SDC=\angle SIB=\angle I_ACB=\angle ETS$ so $ SDTE$ is cyclic and $\{ P, Q\}=\{ S, T\}$ note that $\angle PMQ=2(\angle PEQ)=2(\angle PEB+\angle QEB)=2(\angle AI_AB+\angle AI_AC)=2\angle BI_AC=180-\angle BAC$. DONE

Another Solution: Again $AB<AC$. It is enough to show that $PM \parallel AC$, then analogously $QM \parallel AB$, and we get $\angle PMQ = \angle PMD+ \angle DMQ = \gamma + \beta = 180^\circ - \alpha$. Let $K=AI \cap BC$. We want that $\frac{KM}{KC}=\frac{PM}{AC}$. We easily calculate $KM=\frac{ab}{b+c}-\frac{a}{2}$, $KC=\frac{ab}{b+c}$ and $PM=DM=\frac{b-c}{2}$. Now it ist straightforward that the assertion is true.

Dear Mathlinkers, the circle W is also a Morel's circle in a general case... See http://jl.ayme.pagesperso-orange.fr/Docs/Feuerbach.pdf p. 3 and 5 Sincerely Jean-Louis

This is essentially the same as USA TST 2015 #1 by Evan Chen.