Suppose $P$ is a quintic polynomial with real coefficients with $P(0)=2$ and $P(1)=3$ such that $|z|=1$ whenever $z$ is a complex number satisfying $P(z) = 0$.  What is the smallest possible value of $P(2)$ over all such polynomials $P$?
1. **Identify the roots and form of the polynomial:**
   Given that \(P\) is a quintic polynomial with real coefficients and that all roots \(z\) of \(P(z) = 0\) satisfy \(|z| = 1\), we can write the roots as \(-1, a, \overline{a}, b, \overline{b}\), where \(a\) and \(b\) are complex numbers on the unit circle (i.e., \(|a| = 1\) and \(|b| = 1\)).

2. **Form the polynomial:**
   Since \(P(0) = 2\), we can write \(P(x)\) in the form:
   \[
   P(x) = k(x + 1)(x - a)(x - \overline{a})(x - b)(x - \overline{b})
   \]
   where \(k\) is a constant to be determined.

3. **Determine the constant \(k\):**
   Using \(P(0) = 2\):
   \[
   P(0) = k(1)(-a)(-\overline{a})(-b)(-\overline{b}) = k \cdot 1 \cdot |a|^2 \cdot |b|^2 = k \cdot 1 \cdot 1 \cdot 1 = k
   \]
   Therefore, \(k = 2\).

4. **Simplify the polynomial:**
   \[
   P(x) = 2(x + 1)(x - a)(x - \overline{a})(x - b)(x - \overline{b})
   \]

5. **Use the given \(P(1) = 3\):**
   \[
   P(1) = 2(1 + 1)(1 - a)(1 - \overline{a})(1 - b)(1 - \overline{b}) = 3
   \]
   Simplifying:
   \[
   2 \cdot 2 \cdot |1 - a|^2 \cdot |1 - b|^2 = 3
   \]
   \[
   4 |1 - a|^2 |1 - b|^2 = 3
   \]
   \[
   |1 - a|^2 |1 - b|^2 = \frac{3}{4}
   \]

6. **Minimize \(P(2)\):**
   \[
   P(2) = 2(2 + 1)(2 - a)(2 - \overline{a})(2 - b)(2 - \overline{b})
   \]
   \[
   P(2) = 2 \cdot 3 \cdot |2 - a|^2 \cdot |2 - b|^2
   \]
   We need to minimize \(6 |2 - a|^2 |2 - b|^2\).

7. **Use the constraint \(|1 - a|^2 |1 - b|^2 = \frac{3}{4}\):**
   Since \(a\) and \(b\) lie on the unit circle, we can write \(a = e^{i\theta}\) and \(b = e^{i\phi}\). Then:
   \[
   |1 - a|^2 = (1 - \cos\theta)^2 + \sin^2\theta = 2(1 - \cos\theta)
   \]
   \[
   |1 - b|^2 = (1 - \cos\phi)^2 + \sin^2\phi = 2(1 - \cos\phi)
   \]
   \[
   4(1 - \cos\theta)(1 - \cos\phi) = \frac{3}{4}
   \]
   \[
   (1 - \cos\theta)(1 - \cos\phi) = \frac{3}{16}
   \]

8. **Minimize \(|2 - a|^2\) and \(|2 - b|^2\):**
   \[
   |2 - a|^2 = (2 - \cos\theta)^2 + \sin^2\theta = 4 - 4\cos\theta + 1 = 5 - 4\cos\theta
   \]
   \[
   |2 - b|^2 = (2 - \cos\phi)^2 + \sin^2\phi = 4 - 4\cos\phi + 1 = 5 - 4\cos\phi
   \]
   \[
   |2 - a|^2 |2 - b|^2 = (5 - 4\cos\theta)(5 - 4\cos\phi)
   \]

9. **Use the constraint to find the minimum:**
   \[
   (5 - 4\cos\theta)(5 - 4\cos\phi) = 25 - 20(\cos\theta + \cos\phi) + 16\cos\theta\cos\phi
   \]
   Using \((1 - \cos\theta)(1 - \cos\phi) = \frac{3}{16}\):
   \[
   \cos\theta + \cos\phi = 2 - \frac{3}{8} = \frac{13}{8}
   \]
   \[
   \cos\theta\cos\phi = \frac{1}{2}(\cos(\theta + \phi) + \cos(\theta - \phi))
   \]
   The minimum value occurs when \(\cos\theta = \cos\phi = \frac{1}{2}\):
   \[
   (5 - 4 \cdot \frac{1}{2})(5 - 4 \cdot \frac{1}{2}) = (5 - 2)(5 - 2) = 3 \cdot 3 = 9
   \]

10. **Calculate \(P(2)\):**
    \[
    P(2) = 6 \cdot 9 = 54
    \]

The final answer is \(\boxed{54}\).