Solution: First of all note that the discriminant of the polynomial is given by \begin{align*} a^2 - 4p^{11}. \end{align*} From the given condition, we have \[ a^2 \leq 4p^{11} \implies a^2 - 4p^{11} \leq 0. \] If $a^2 - 4p^{11}=0$, then \[ a^2 = 4p^{11} \implies a = \pm \sqrt{4p^{11}} = \pm 2p^{5}\sqrt{p}. \]

Since $p$ is a prime number, $\sqrt{p} $ must be an irrational number and hence $a$ is an irrational number, which is a contradiction to the fact that $a\in \mathbb{Z} \setminus \{0\}$. Therefore, $a^2 < 4p^{11}$, and hence the roots of this quadratic equation will be distinct and imaginary roots. Thus, option (D) is wrong and option (C) is correct.

Now we will analyze the other two options. Recall that roots of the quadratic equation $az^2 + bz + c$ is given by \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] We have $a^2 - 4p^{11} \leq 0$, so the roots will be \begin{align*} z = \frac{-a \pm \sqrt{a^2 - 4p^{11}} }{2} = \frac{-a \pm \iota \sqrt{4p^{11}-a^2} }{2}. \end{align*} The real part and the imaginary parts are \begin{align*} \text{real part } & = \frac{-a}{2} \text{ and } \\ \text{imaginary part } & = \pm \frac{\sqrt{4p^{11}-a^2}}{2}. \end{align*} Note that $a\neq 0$, so real part is non zero and hence option (A) is also false.

Finally, if option (B) is correct, then \begin{align*} -\frac{a}{2} = \pm \frac{\sqrt{4p^{11}-a^2} }{2} & \implies a^2 = 4p^{11} - a^2 \\ & \implies 2a^2 = 4p^{11} \\ & \implies a^2 = p^{11} \\ & \implies a = \pm 2\sqrt{p^{11} } = \pm p^5\sqrt{p}, \end{align*} which is a contradiction as $p$ is a prime number and $a$ is an integer. Therefore, option (B) is also false. Hence the only option which is correct is option (C).