[THOUGHT] We set up the system analytically and will now solve for $a$ using Python, with $p=1$ and the derived conditions:

1. From geometry of foci: $a^2 - b^2 = 1$.
2. From intersection point with parameter $t$ such that $\angle P_1 A P_2 = 60^\circ$, we got $t^2 = 3$.
3. The intersection point $(x,y) = (t^2, 2t)$ lies on the ellipse:

Substitute $t^2 = 3$ and $b^2 = a^2 - 1$, solve for $a$, then compute $\left\lfloor a^3 \right\rfloor$ which equals $\left\lfloor \frac{a^3}{p^3} \right\rfloor$ since $p=1$. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

a = sp.symbols('a', positive=True)

Given relationships with p=1

t2 = 3 t4 = t2**2

b^2 = a^2 - 1

b2 = a**2 - 1

Ellipse intersection condition: t^4/a^2 + 4 t^2 / b^2 = 1

eq = sp.Eq(t4/a**2 + 4*t2/b2, 1)

Solve for a^2 first for stability, then take sqrt

A2 = sp.symbols('A2', positive=True) eq_A2 = eq.subs(a**2, A2)

solutions_A2 = sp.solve(eq_A2, A2) solutions_A2 print("Solutions for a^2:", solutions_A2)

Filter real positive solutions for a^2

real_pos_A2 = [sol for sol in solutions_A2 if sol.is_real and sol > 0] print("Real positive a^2 solutions:", real_pos_A2)

Take corresponding a > 0

a_vals = [sp.sqrt(sol) for sol in real_pos_A2] print("a values:", a_vals)

Compute floor(a^3)

for aval in a_vals: a_num = sp.N(aval) val = sp.floor(a_num**3) print("a approx:", a_num, "floor(a^3):", val)
[/PYTHON]

[THOUGHT] From the constraints of the ellipse (with $a>b>0$ and $a^2-b^2=1$), only the larger positive solution for $a$ is valid: $a \approx 4.645751311$. The corresponding Python output gives $\lfloor a^3 \rfloor = 100$. Since $p=1$, this equals $\left\lfloor \frac{a^3}{p^3} \right\rfloor$.

$\boxed{100}$