If q is the average sewage flow from a city of population P, the maximum sewage flow A. $${\text{Q}} = \left( {\frac{{4 + \sqrt {\text{P}} }}{{18 + \sqrt {\text{P}} }}} \right){\text{q}}$$ B. $${\text{Q}} = \left( {\frac{{18 + {\text{P}}}}{{4 + \sqrt {\text{P}} }}} \right){\text{q}}$$ C. $${\text{Q}} = \left( {\frac{{18 + \sqrt {\text{P}} }}{{4 + \sqrt {\text{P}} }}} \right){\text{q}}$$ D. $${\text{Q}} = \left( {\frac{{5 + \sqrt {\text{P}} }}{{15 + \sqrt {\text{P}} }}} \right){\text{q}}$$

The correct answer is $\boxed{\text{Q} = \left( {\frac{{4 + \sqrt {\text{P}} }}{{18 + \sqrt {\text{P}} }}} \right){\text{q}}}$.

The average sewage flow from a city of population $P$ is $q$. The maximum sewage flow is the average flow plus a factor that accounts for the fact that sewage flow is not constant throughout the day. This factor is called the peak factor, and it is typically between 1.5 and 2.

The maximum sewage flow is therefore given by

$$\text{Q} = q(1 + \text{peak factor})$$

The peak factor is a function of the city’s population, and it can be estimated using the following equation:

$$\text{peak factor} = \frac{18 + \sqrt{P}}{4 + \sqrt{P}}$$

Substituting this into the equation for the maximum sewage flow, we get

$$\text{Q} = q\left(1 + \frac{18 + \sqrt{P}}{4 + \sqrt{P}}\right) = \left( {\frac{{4 + \sqrt {\text{P}} }}{{18 + \sqrt {\text{P}} }}} \right){\text{q}}$$

This is the equation for the maximum sewage flow, and it is the correct answer to the question.

The other options are incorrect because they do not account for the peak factor. Option A does not include the peak factor at all, while options B and C include the peak factor but use the wrong formula for it.