On a semicircle of diameter $AB$ and center $C$, consider vari­able points $M$ and $N$ such that $MC \perp NC$. The circumcircle of triangle $MNC$ intersects $AB$ for the second time at $P$. Prove that $\frac{|PM-PN|}{PC}$  constant and find its value. 
1. **Define the coordinates of points:**
   - Let the diameter \( AB \) of the semicircle be \( 2a \), with \( A(-a,0) \), \( C(0,0) \), and \( B(a,0) \).
   - Let \( M \) and \( N \) be points on the semicircle such that \( MC \perp NC \). We can parameterize these points as:
     \[
     M(a \cos \alpha, a \sin \alpha) \quad \text{and} \quad N(-a \sin \alpha, a \cos \alpha)
     \]
     where \( \alpha \) is a variable angle.

2. **Find the coordinates of point \( P \):**
   - The circumcircle of \( \triangle MNC \) intersects \( AB \) at point \( P \). Since \( P \) lies on \( AB \), its y-coordinate is 0. Let \( P \) be \( (x, 0) \).
   - The circumcircle of \( \triangle MNC \) has the equation:
     \[
     (x - a \cos \alpha)^2 + y^2 = a^2 \sin^2 \alpha
     \]
     \[
     (x + a \sin \alpha)^2 + y^2 = a^2 \cos^2 \alpha
     \]
   - Since \( P \) lies on the x-axis, we substitute \( y = 0 \) into the equations:
     \[
     (x - a \cos \alpha)^2 = a^2 \sin^2 \alpha
     \]
     \[
     (x + a \sin \alpha)^2 = a^2 \cos^2 \alpha
     \]
   - Solving these equations, we find:
     \[
     x = a (\cos \alpha - \sin \alpha)
     \]
     Thus, \( P \) is at \( (a (\cos \alpha - \sin \alpha), 0) \).

3. **Calculate distances \( PM \) and \( PN \):**
   - Distance \( PM \):
     \[
     PM = \sqrt{(a (\cos \alpha - \sin \alpha) - a \cos \alpha)^2 + (0 - a \sin \alpha)^2}
     \]
     \[
     PM = \sqrt{(a (- \sin \alpha))^2 + (a \sin \alpha)^2}
     \]
     \[
     PM = \sqrt{a^2 \sin^2 \alpha + a^2 \sin^2 \alpha}
     \]
     \[
     PM = a \sqrt{2} \sin \alpha
     \]
   - Distance \( PN \):
     \[
     PN = \sqrt{(a (\cos \alpha - \sin \alpha) + a \sin \alpha)^2 + (0 - a \cos \alpha)^2}
     \]
     \[
     PN = \sqrt{(a \cos \alpha)^2 + (a \cos \alpha)^2}
     \]
     \[
     PN = a \sqrt{2} \cos \alpha
     \]

4. **Calculate \( PC \):**
   - Distance \( PC \):
     \[
     PC = |a (\cos \alpha - \sin \alpha)|
     \]

5. **Find \( \frac{|PM - PN|}{PC} \):**
   - Calculate \( |PM - PN| \):
     \[
     |PM - PN| = |a \sqrt{2} \sin \alpha - a \sqrt{2} \cos \alpha|
     \]
     \[
     |PM - PN| = a \sqrt{2} |\sin \alpha - \cos \alpha|
     \]
   - Calculate \( \frac{|PM - PN|}{PC} \):
     \[
     \frac{|PM - PN|}{PC} = \frac{a \sqrt{2} |\sin \alpha - \cos \alpha|}{a |\cos \alpha - \sin \alpha|}
     \]
     \[
     \frac{|PM - PN|}{PC} = \sqrt{2}
     \]

Thus, the value of \( \frac{|PM - PN|}{PC} \) is constant and equal to \( \sqrt{2} \).

The final answer is \( \boxed{ \sqrt{2} } \).