 Let $p$ be a real constant. The parabola $y^2=-4px$ rolls without slipping around the parabola $y^2=4px$. Find the equation of the locus of the vertex of the rolling parabola.
1. **Identify the given parabolas and their equations:**
   - The fixed parabola is \( y^2 = 4px \).
   - The rolling parabola is \( y^2 = -4px \).

2. **Determine a point on the fixed parabola:**
   - Let \( A \) be a point on the fixed parabola \( y^2 = 4px \). We can parameterize this point as \( A(p\lambda^2, 2p\lambda) \).

3. **Find the equation of the tangent line at point \( A \):**
   - The slope of the tangent line at \( A \) can be found by differentiating \( y^2 = 4px \):
     \[
     2y \frac{dy}{dx} = 4p \implies \frac{dy}{dx} = \frac{2p}{y}
     \]
   - At \( A(p\lambda^2, 2p\lambda) \), the slope is:
     \[
     \frac{dy}{dx} = \frac{2p}{2p\lambda} = \frac{1}{\lambda}
     \]
   - The equation of the tangent line at \( A \) is:
     \[
     y - 2p\lambda = \frac{1}{\lambda}(x - p\lambda^2) \implies y\lambda = x + p\lambda^2
     \]

4. **Find the equation of the line through the origin perpendicular to the tangent line:**
   - The slope of the line perpendicular to the tangent line \( y\lambda = x + p\lambda^2 \) is \( -\lambda \).
   - The equation of this line through the origin is:
     \[
     y = -\lambda x
     \]

5. **Find the intersection point \( S \) of the tangent line and the perpendicular line:**
   - Substitute \( y = -\lambda x \) into the tangent line equation \( y\lambda = x + p\lambda^2 \):
     \[
     (-\lambda x)\lambda = x + p\lambda^2 \implies -\lambda^2 x = x + p\lambda^2 \implies x(-\lambda^2 - 1) = p\lambda^2 \implies x = \frac{-p\lambda^2}{1 + \lambda^2}
     \]
   - Substitute \( x = \frac{-p\lambda^2}{1 + \lambda^2} \) back into \( y = -\lambda x \):
     \[
     y = -\lambda \left( \frac{-p\lambda^2}{1 + \lambda^2} \right) = \frac{p\lambda^3}{1 + \lambda^2}
     \]
   - Thus, the coordinates of \( S \) are:
     \[
     S \left( \frac{-p\lambda^2}{1 + \lambda^2}, \frac{p\lambda^3}{1 + \lambda^2} \right)
     \]

6. **Find the vertex \( B' \) of the rolling parabola:**
   - The vertex of the rolling parabola is the mirror image of the fixed parabola's vertex with respect to the tangent line.
   - The coordinates of \( B' \) are:
     \[
     B' \left( \frac{-2p\lambda^2}{1 + \lambda^2}, \frac{2p\lambda^3}{1 + \lambda^2} \right)
     \]

7. **Eliminate the parameter \( \lambda \) to find the locus of \( B' \):**
   - Let \( x = \frac{-2p\lambda^2}{1 + \lambda^2} \) and \( y = \frac{2p\lambda^3}{1 + \lambda^2} \).
   - From \( x = \frac{-2p\lambda^2}{1 + \lambda^2} \), we get:
     \[
     \lambda^2 = \frac{-x(1 + \lambda^2)}{2p} \implies \lambda^2 (1 + \frac{x}{2p}) = -\frac{x}{2p} \implies \lambda^2 = \frac{-x}{2p + x}
     \]
   - Substitute \( \lambda^2 \) into \( y = \frac{2p\lambda^3}{1 + \lambda^2} \):
     \[
     y = \frac{2p \left( \frac{-x}{2p + x} \right)^{3/2}}{1 + \frac{-x}{2p + x}} = \frac{2p \left( \frac{-x}{2p + x} \right)^{3/2}}{\frac{2p}{2p + x}} = \frac{2p \left( \frac{-x}{2p + x} \right)^{3/2} (2p + x)}{2p}
     \]
   - Simplify to get the final equation:
     \[
     x(x^2 + y^2) + 2py^2 = 0
     \]

The final answer is \( \boxed{ x(x^2 + y^2) + 2py^2 = 0 } \)