Let the real parameter $p$ be such that the system $\begin{cases} p(x^2 - y^2) = (p^2- 1)xy \\ |x - 1|+ |y| = 1 \end{cases}$ has at least three different real solutions. Find $p$ and solve the system for that $p$.
1. **Analyze the given system of equations:**
   \[
   \begin{cases}
   p(x^2 - y^2) = (p^2 - 1)xy \\
   |x - 1| + |y| = 1
   \end{cases}
   \]

2. **Consider the first equation:**
   \[
   p(x^2 - y^2) = (p^2 - 1)xy
   \]
   If \( p \neq 0 \), we can rewrite it as:
   \[
   x^2 - y^2 = \frac{p^2 - 1}{p}xy
   \]
   Rearrange to:
   \[
   x^2 - \frac{p^2 - 1}{p}xy - y^2 = 0
   \]
   This is a quadratic equation in \( y \):
   \[
   y^2 - \frac{p^2 - 1}{p}xy - x^2 = 0
   \]
   Solving for \( y \):
   \[
   y = \frac{\frac{p^2 - 1}{p}x \pm \sqrt{\left(\frac{p^2 - 1}{p}x\right)^2 + 4x^2}}{2}
   \]
   Simplify the discriminant:
   \[
   \left(\frac{p^2 - 1}{p}x\right)^2 + 4x^2 = \frac{(p^2 - 1)^2}{p^2}x^2 + 4x^2 = \left(\frac{(p^2 - 1)^2 + 4p^2}{p^2}\right)x^2 = \frac{p^4 - 2p^2 + 1 + 4p^2}{p^2}x^2 = \frac{p^4 + 2p^2 + 1}{p^2}x^2 = \left(\frac{p^2 + 1}{p}\right)^2 x^2
   \]
   Thus:
   \[
   y = \frac{\frac{p^2 - 1}{p}x \pm \frac{p^2 + 1}{p}x}{2} = \frac{(p^2 - 1)x \pm (p^2 + 1)x}{2p} = \frac{(p^2 - 1 \pm (p^2 + 1))x}{2p}
   \]
   This gives two solutions:
   \[
   y = \frac{2p^2 x}{2p} = px \quad \text{or} \quad y = \frac{-2x}{2p} = -\frac{x}{p}
   \]
   Therefore, the solutions are:
   \[
   y = px \quad \text{or} \quad y = -\frac{x}{p}
   \]

3. **Consider the second equation:**
   \[
   |x - 1| + |y| = 1
   \]
   We need to find the intersections of the lines \( y = px \) and \( y = -\frac{x}{p} \) with the square defined by \( |x - 1| + |y| = 1 \).

4. **Check the intersections for \( y = px \):**
   \[
   |x - 1| + |px| = 1
   \]
   - For \( x \geq 1 \):
     \[
     x - 1 + px = 1 \implies x(1 + p) = 2 \implies x = \frac{2}{1 + p}
     \]
     \[
     y = p \cdot \frac{2}{1 + p} = \frac{2p}{1 + p}
     \]
   - For \( x < 1 \):
     \[
     1 - x + px = 1 \implies x(p - 1) = 0 \implies x = 0
     \]
     \[
     y = 0
     \]

5. **Check the intersections for \( y = -\frac{x}{p} \):**
   \[
   |x - 1| + \left| -\frac{x}{p} \right| = 1
   \]
   - For \( x \geq 1 \):
     \[
     x - 1 + \frac{x}{p} = 1 \implies x(1 + \frac{1}{p}) = 2 \implies x = \frac{2p}{p + 1}
     \]
     \[
     y = -\frac{2p}{p(p + 1)} = -\frac{2}{p + 1}
     \]
   - For \( x < 1 \):
     \[
     1 - x + \frac{x}{p} = 1 \implies x(\frac{1}{p} - 1) = 0 \implies x = 0
     \]
     \[
     y = 0
     \]

6. **Determine the value of \( p \) such that there are at least three different solutions:**
   - For \( p = 1 \):
     \[
     y = x \quad \text{and} \quad y = -x
     \]
     The intersections are:
     \[
     (0, 0), (1, 1), (1, -1)
     \]
     This gives three different solutions.

   - For \( p = -1 \):
     \[
     y = -x \quad \text{and} \quad y = x
     \]
     The intersections are:
     \[
     (0, 0), (1, -1), (1, 1)
     \]
     This also gives three different solutions.

Therefore, the values of \( p \) that satisfy the condition are \( p = \pm 1 \).

The final answer is \( \boxed{ p = \pm 1 } \)