For each polynomial $P(x)$, define $$P_1(x)=P(x), \forall x \in \mathbb{R},$$ $$P_2(x)=P(P_1(x)),  \forall x \in \mathbb{R},$$ $$...$$ $$P_{2024}(x)=P(P_{2023}(x)), \forall x \in \mathbb{R}.$$ Let $a>2$ be a real number. Is there a polynomial $P$ with real coefficients such that for all $t \in (-a, a)$, the equation $P_{2024}(x)=t$ has $2^{2024}$ distinct real roots?
1. **Define the sequence of polynomials:**
   \[
   P_1(x) = P(x), \quad P_2(x) = P(P_1(x)), \quad \ldots, \quad P_{2024}(x) = P(P_{2023}(x))
   \]
   We need to find a polynomial \( P \) such that for all \( t \in (-a, a) \), the equation \( P_{2024}(x) = t \) has \( 2^{2024} \) distinct real roots.

2. **Choose a specific polynomial:**
   Consider the polynomial \( P(x) = Cx(x-1) \) where \( C \geq 4a \) and \( C+2 > 2\sqrt{1 + \frac{4a}{C}} \). This choice ensures that the polynomial has the desired properties.

3. **Base case for induction:**
   For \( n = 1 \), the equation \( P_1(x) = t \) becomes:
   \[
   Cx(x-1) = t
   \]
   The discriminant of this quadratic equation is:
   \[
   \Delta = 1 + \frac{4t}{C} = \frac{C + 4t}{C}
   \]
   Since \( C \geq 4a \) and \( t \in (-a, a) \), we have \( C + 4t > 0 \). Therefore, the equation has two distinct real roots:
   \[
   x_1 = \frac{1 - \sqrt{1 + \frac{4t}{C}}}{2}, \quad x_2 = \frac{1 + \sqrt{1 + \frac{4t}{C}}}{2}
   \]

4. **Inductive step:**
   Assume that for \( P_n(x) = t \), the equation has \( 2^n \) distinct real roots. We need to show that \( P_{n+1}(x) = t \) has \( 2^{n+1} \) distinct real roots.

5. **Define the sequence \( (u_n) \):**
   Let \( u_0 = t \) and for each \( 1 \leq i \leq n-1 \), let \( u_i \) denote the greatest root among the \( 2^{i-1} \) roots of the equation \( P_i(x) = t \). It is trivial that \( u_n > 0 \) for all \( n > 1 \).

6. **Roots of the equation \( Cx(x-1) = k \):**
   For some real number \( k > -\frac{C}{4} \), the equation \( Cx(x-1) = k \) has two solutions:
   \[
   x = \frac{1 + \sqrt{1 + \frac{4k}{C}}}{2}, \quad x = \frac{1 - \sqrt{1 + \frac{4k}{C}}}{2}
   \]
   When \( k \) increases, the larger root increases and the smaller root decreases.

7. **Sequence \( (u_n) \) properties:**
   \[
   u_{n+1} = \frac{1 + \sqrt{1 + \frac{4u_n}{C}}}{2}
   \]
   Since \( t < a \) and \( C > 4a \), it follows that:
   \[
   u_1 = \frac{1 + \sqrt{1 + \frac{4t}{C}}}{2} < \frac{1 + \sqrt{1 + \frac{4a}{4a}}}{2} < a
   \]
   By induction, \( u_n < a \) for all positive integers \( n \).

8. **Equation \( Cx(x-1) = u_{n-1} \):**
   The equation has two roots:
   \[
   x_1 = \frac{1 + \sqrt{1 + \frac{4u_{n-1}}{C}}}{2}, \quad x_2 = \frac{1 - \sqrt{1 + \frac{4u_{n-1}}{C}}}{2}
   \]
   Since \( u_{n-1} \) is the largest root among the \( 2^{n-1} \) roots, \( x_1 \) and \( x_2 \) are the largest and smallest roots among the \( 2^n \) roots of \( P_n(x) = t \).

9. **Discriminant of \( Cx(x-1) = x_2 \):**
   \[
   1 + \frac{4x_2}{C} = \frac{1}{C} \left( C + 2 \left( 1 - \sqrt{1 + \frac{4u_{n-1}}{C}} \right) \right) > \frac{1}{C} \left( C + 2 - 2\sqrt{1 + \frac{4a}{C}} \right) > 0
   \]
   This means the equation \( Cx(x-1) = x_2 \) has two distinct roots. Therefore, for each of the \( 2^n \) roots \( y \) of \( P_n(x) = t \), the equation \( Cx(x-1) = y \) has two distinct roots, giving a total of \( 2 \cdot 2^n = 2^{n+1} \) roots.

10. **Conclusion:**
    By induction, the equation \( P_{2024}(x) = t \) has \( 2^{2024} \) distinct real roots for all \( t \in (-a, a) \).

The final answer is \(\boxed{\text{Yes}}\)