Find the number of primes$ p$, such that $ x^{3} \minus{}5x^{2} \minus{}22x\plus{}56\equiv 0\, \, \left(mod\, p\right)$ has no three distinct integer roots in $ \left[0,\left. p\right)\right.$.

$\textbf{(A)}\ 1  \qquad\textbf{(B)}\ 2  \qquad\textbf{(C)}\ 3  \qquad\textbf{(D)}\ 4  \qquad\textbf{(E)}\ \text{None}$
1. We start with the polynomial \( x^3 - 5x^2 - 22x + 56 \). We need to find the number of primes \( p \) such that the polynomial has no three distinct integer roots in the range \([0, p)\).

2. First, we factorize the polynomial:
   \[
   x^3 - 5x^2 - 22x + 56 = (x - 2)(x - 7)(x + 4)
   \]
   This factorization can be verified by expanding the product:
   \[
   (x - 2)(x - 7)(x + 4) = x^3 - 7x^2 + 4x - 2x^2 + 14x - 8x + 28 = x^3 - 5x^2 - 22x + 56
   \]

3. The roots of the polynomial are \( x = 2, 7, -4 \). We need to check when these roots are not distinct modulo \( p \).

4. Consider the congruences:
   \[
   2 \equiv 7 \pmod{p} \implies 2 - 7 \equiv 0 \pmod{p} \implies -5 \equiv 0 \pmod{p} \implies p = 5
   \]

5. Next, consider:
   \[
   2 \equiv -4 \pmod{p} \implies 2 + 4 \equiv 0 \pmod{p} \implies 6 \equiv 0 \pmod{p} \implies p = 2 \text{ or } p = 3
   \]

6. Finally, consider:
   \[
   7 \equiv -4 \pmod{p} \implies 7 + 4 \equiv 0 \pmod{p} \implies 11 \equiv 0 \pmod{p} \implies p = 11
   \]

7. We have found the primes \( p = 2, 3, 5, 11 \). These are the primes for which the polynomial \( x^3 - 5x^2 - 22x + 56 \equiv 0 \pmod{p} \) does not have three distinct integer roots in the range \([0, p)\).

8. Therefore, the number of such primes is 4.

The final answer is \(\boxed{4}\).