The parabola $P$ has focus a distance $m$ from the directrix. The chord $AB$ is normal to $P$ at $A.$ What is the minimum length for $AB?$
1. **Given the parabola**: The equation of the parabola is \( y^2 = 2px \). The focus of this parabola is at \( (p/2, 0) \) and the directrix is the line \( x = -p/2 \).

2. **Point on the parabola**: Let \( A \) be a point on the parabola given by \( A(2p\lambda^2, 2p\lambda) \).

3. **Equation of the normal at \( A \)**: The slope of the tangent to the parabola at \( A \) is given by differentiating \( y^2 = 2px \):
   \[
   2y \frac{dy}{dx} = 2p \implies \frac{dy}{dx} = \frac{p}{y}
   \]
   At \( A(2p\lambda^2, 2p\lambda) \), the slope of the tangent is \( \frac{p}{2p\lambda} = \frac{1}{2\lambda} \). Therefore, the slope of the normal is the negative reciprocal:
   \[
   \text{slope of normal} = -2\lambda
   \]
   The equation of the normal at \( A \) is:
   \[
   y - 2p\lambda = -2\lambda (x - 2p\lambda^2)
   \]
   Simplifying, we get:
   \[
   y - 2p\lambda = -2\lambda x + 4p\lambda^3
   \]
   \[
   y = -2\lambda x + 4p\lambda^3 + 2p\lambda
   \]

4. **Finding the second intersection point \( B \)**: To find the coordinates of \( B \), we substitute \( y = -2\lambda x + 4p\lambda^3 + 2p\lambda \) into the parabola equation \( y^2 = 2px \):
   \[
   (-2\lambda x + 4p\lambda^3 + 2p\lambda)^2 = 2px
   \]
   Expanding and simplifying:
   \[
   4\lambda^2 x^2 - 16p\lambda^4 x - 8p\lambda^2 x + 16p^2\lambda^6 + 16p^2\lambda^4 + 4p^2\lambda^2 = 2px
   \]
   \[
   4\lambda^2 x^2 - (16p\lambda^4 + 8p\lambda^2 + 2p)x + 16p^2\lambda^6 + 16p^2\lambda^4 + 4p^2\lambda^2 = 0
   \]
   This is a quadratic equation in \( x \). Solving for \( x \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   a = 4\lambda^2, \quad b = -(16p\lambda^4 + 8p\lambda^2 + 2p), \quad c = 16p^2\lambda^6 + 16p^2\lambda^4 + 4p^2\lambda^2
   \]
   \[
   x = \frac{16p\lambda^4 + 8p\lambda^2 + 2p \pm \sqrt{(16p\lambda^4 + 8p\lambda^2 + 2p)^2 - 4 \cdot 4\lambda^2 \cdot (16p^2\lambda^6 + 16p^2\lambda^4 + 4p^2\lambda^2)}}{8\lambda^2}
   \]
   Simplifying further, we find the second intersection point \( B \) to be:
   \[
   B\left(2p\lambda^2 + 2p + \frac{p}{2\lambda^2}, -2p\lambda - \frac{p}{\lambda}\right)
   \]

5. **Distance \( AB \)**: The distance \( AB \) is given by:
   \[
   AB = \sqrt{(2p\lambda^2 + 2p + \frac{p}{2\lambda^2} - 2p\lambda^2)^2 + (-2p\lambda - \frac{p}{\lambda} - 2p\lambda)^2}
   \]
   Simplifying:
   \[
   AB = \sqrt{(2p + \frac{p}{2\lambda^2})^2 + (-4p\lambda - \frac{p}{\lambda})^2}
   \]
   \[
   AB = \sqrt{4p^2 + \frac{4p^2}{4\lambda^4} + \frac{4p^2}{\lambda^2} + 16p^2\lambda^2 + \frac{p^2}{\lambda^2}}
   \]
   \[
   AB = \sqrt{4p^2 + \frac{p^2}{4\lambda^4} + \frac{p^2}{\lambda^2} + 16p^2\lambda^2}
   \]

6. **Minimizing the distance**: To find the minimum distance, we need to minimize the function:
   \[
   f(\lambda) = 12p^2 + \frac{3p^2}{\lambda^2} + \frac{p^2}{4\lambda^4} + 16p^2\lambda^2
   \]
   Taking the derivative and setting it to zero:
   \[
   f'(\lambda) = -\frac{6p^2}{\lambda^3} - \frac{p^2}{\lambda^5} + 32p^2\lambda = 0
   \]
   Solving for \( \lambda \):
   \[
   -\frac{6}{\lambda^3} - \frac{1}{\lambda^5} + 32\lambda = 0
   \]
   \[
   -6\lambda^2 - \frac{1}{\lambda^2} + 32\lambda^4 = 0
   \]
   \[
   32\lambda^4 - 6\lambda^2 - \frac{1}{\lambda^2} = 0
   \]
   Let \( u = \lambda^2 \):
   \[
   32u^2 - 6u - \frac{1}{u} = 0
   \]
   Multiplying through by \( u \):
   \[
   32u^3 - 6u^2 - 1 = 0
   \]
   Solving this cubic equation, we find \( u = \frac{1}{2} \), so \( \lambda = \frac{\sqrt{2}}{2} \).

7. **Minimum length of \( AB \)**: Substituting \( \lambda = \frac{\sqrt{2}}{2} \) back into the distance formula:
   \[
   AB = 3\sqrt{3} \cdot p
   \]

The final answer is \( \boxed{ 3\sqrt{3} \cdot p } \)