jmc

We claim that the product of the non-real zeros is the smallest.

(A) The value of $P(-1)$ is greater than 4.

(B) Since the leading coefficient in $P(x)$ is 1, the product of the zeros of $P$ is $d=P(0),$ which is greater than 4.

(D) The sum of the coefficient of $P(x)$ is $P(1),$ which is greater than 2.

(E) The quartic $P(x)$ has a real root between 1 and 2, and it also has a root between 3 and 4. If there were any more real roots, then the quartic equation $P(x)=5$ would have more than four roots, which is impossible, so these two real roots are the only real roots. The sum of these real roots is greater than 4.

(C) The product of all the zeros is $d=P(0),$ which is less than 6. The product of the real zeros is greater than 3, so the product the non-real zeros must be less than $\frac{6}{3}=2.$

Thus, the answer is $\boxed{\text{C}}.$