Given triangle $ABC$. Point $M$ is the projection of vertex $B$ to bisector of angle $C$. $K$ is the touching point of the incircle with side $BC$. Find angle $\angle  MKB$ if $\angle BAC = \alpha$

(V.Protasov) 
1. **Understanding the Problem:**
   - We are given a triangle \(ABC\).
   - Point \(M\) is the projection of vertex \(B\) onto the angle bisector of \(\angle C\).
   - Point \(K\) is the point where the incircle of \(\triangle ABC\) touches side \(BC\).
   - We need to find the angle \(\angle MKB\) given that \(\angle BAC = \alpha\).

2. **Using Sharigin Problem 255:**
   - According to Sharigin problem 255, if points \(L\) and \(N\) are the points where the incircle touches sides \(AB\) and \(AC\) respectively, then points \(N\), \(L\), and \(M\) are collinear.

3. **Collinearity and Angle Calculation:**
   - Since \(N\), \(L\), and \(M\) are collinear, we can use this collinearity to find the required angle.
   - The angle \(\angle MKB\) is equal to \(\angle MLB\) because \(M\) lies on the angle bisector of \(\angle C\) and \(K\) is the tangency point on \(BC\).

4. **Using the Property of the Incircle:**
   - The angle \(\angle ALN\) is known to be \(\frac{\pi - \alpha}{2}\) because the incircle tangency points create equal angles with the angle bisectors.

5. **Conclusion:**
   - Therefore, \(\angle MKB = \angle MLB = \angle ALN = \frac{\pi - \alpha}{2}\).

\[
\boxed{\frac{\pi - \alpha}{2}}
\]