The correct answer is $\boxed{{1 \over {\left( {0.5 – j0.25} \right)}}}$.
Let $x[n]$ be a sequence with even symmetry, i.e., $x[n] = x[-n]$. The z-transform of $x[n]$ is given by
$$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} = \sum_{n=-\infty}^{\infty} x[-n] z^{n} = \sum_{n=0}^{\infty} x[n] z^{n}$$
Therefore, the poles of $X(z)$ are the complex numbers $z$ such that $z^{-1}$ is a zero of $X(z)$.
In this case, we are given that $0.5 + j0.25$ is a zero of $X(z)$. Therefore, $z^{-1} = 0.5 – j0.25$ is a pole of $X(z)$. The reciprocal of a pole is a zero, so $1 / (0.5 – j0.25)$ is also a zero of $X(z)$.
The other options are not necessarily zeros of $X(z)$. For example, $0.5 + j0.25$ is not a zero of $X(z)$, since $z^{-1} = 0.5 + j0.25$ is not a pole of $X(z)$.