Let $A', B' , C'$ be the projections of a point $M$ inside a triangle $ABC$ onto the sides $BC, CA, AB$, respectively. Deﬁne $p(M ) = \frac{MA'\cdot MB'\cdot MC'}{MA \cdot MB \cdot MC}$ . Find the position of point $M$ that maximizes $p(M )$.
1. Let \( A', B', C' \) be the projections of a point \( M \) inside a triangle \( ABC \) onto the sides \( BC, CA, AB \), respectively. Define \( MA' = d_a \), \( MB' = d_b \), \( MC' = d_c \), \( \angle MAB = \alpha_c \), and \( \angle MAC = \alpha_b \).

2. We need to find the position of point \( M \) that maximizes \( p(M) = \frac{MA' \cdot MB' \cdot MC'}{MA \cdot MB \cdot MC} \).

3. Consider the product \( MB \cdot MC \). Using the Law of Sines in triangle \( MBC \), we have:
   \[
   MB \cdot MC = \frac{d_b d_c}{\sin \alpha_c \sin \alpha_b}
   \]

4. Using the trigonometric identity for the product of sines, we get:
   \[
   \sin \alpha_c \sin \alpha_b = \frac{1}{2} [\cos(\alpha_c - \alpha_b) - \cos(\alpha_c + \alpha_b)]
   \]

5. Since \( \cos(\alpha_c + \alpha_b) \leq 1 \), we have:
   \[
   \cos(\alpha_c - \alpha_b) - \cos(\alpha_c + \alpha_b) \geq \cos(\alpha_c - \alpha_b) - 1
   \]

6. Therefore:
   \[
   MB \cdot MC \geq \frac{2 d_b d_c}{1 - \cos A} = \frac{d_b d_c}{\sin^2 \frac{A}{2}}
   \]

7. Similarly, we can derive:
   \[
   MC \cdot MA \geq \frac{d_c d_a}{\sin^2 \frac{B}{2}}
   \]
   \[
   MA \cdot MB \geq \frac{d_a d_b}{\sin^2 \frac{C}{2}}
   \]

8. Multiplying these inequalities together and taking the square root, we get:
   \[
   \frac{d_a d_b d_c}{MA \cdot MB \cdot MC} \leq \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
   \]

9. Equality holds if and only if \( M \equiv I \), the incenter of \( \triangle ABC \).

The final answer is \( \boxed{ M \equiv I } \), the incenter of \( \triangle ABC \).