The correct answer is $\boxed{\frac{{1 + a{z^{ – 1}}}}{{1 – b{z^{ – 1}}}}}$.
The transfer function of a system is the ratio of its output to its input, expressed in terms of the complex frequency variable $z$. In the case of a first-order system, the transfer function can be written as
$$H(z) = \frac{{1 + a{z^{ – 1}}}}{{1 – b{z^{ – 1}}}}$$
where $a$ is the system’s pole and $b$ is the system’s zero.
In the figure, the system is a first-order low-pass filter with a pole at $z = -a$ and a zero at $z = -b$. Therefore, the transfer function of the system is
$$H(z) = \frac{{1 + a{z^{ – 1}}}}{{1 – b{z^{ – 1}}}}$$
Option A is incorrect because it does not have a zero. Option B is incorrect because it does not have a pole. Option C is incorrect because it has a pole at $z = -a$ and a zero at $z = -b$, which is not the case for the system in the figure. Option D is incorrect because it has a pole at $z = -b$ and a zero at $z = -a$, which is not the case for the system in the figure.