To determine the correct conditions for which a hydraulically efficient rectangular channel delivers maximum discharge, we need to analyze each given condition and understand some basic concepts of open channel flow.

For a rectangular channel to be hydraulically efficient, it should minimize the wetted perimeter for a given cross-sectional area to maximize the hydraulic radius, \(R_h\), which is given by:

\(R_h = \frac{A}{P}\)

where:

• \(A\) is the area of flow.
• \(P\) is the wetted perimeter.

The condition for maximum discharge in a most efficient rectangular channel is when the hydraulic radius is maximized. For a rectangular channel of breadth \(b\) and depth \(d\), the area \(A = b \cdot d\), and the wetted perimeter \(P = b + 2d\). Thus, hydraulic radius \(R_h = \frac{b \cdot d}{b + 2d}\).

Now, let's evaluate each condition:

1. Condition P: Depth of water is equal to half the breadth of the channel.
   When \(d = \frac{b}{2}\), \(R_h = \frac{\left(b \cdot \frac{b}{2}\right)}{b + b} = \frac{\frac{b^2}{2}}{2b} = \frac{b}{4}\). This condition aligns with maximizing hydraulic efficiency.
2. Condition Q: Depth of water is equal to the breadth of the channel.
   This makes the channel square, giving \(R_h = \frac{b \cdot b}{b + 2b} = \frac{b^2}{3b} = \frac{b}{3}\), which is not optimal.
3. Condition R: Depth of water is equal to twice the breadth of the channel.
   Here, \(d = 2b\), giving \(R_h = \frac{b \cdot 2b}{b + 4b} = \frac{2b^2}{5b} = \frac{2b}{5}\), which is not optimal.
4. Condition S: Hydraulic radius is equal to half the depth of water.
   Let \(R_h = \frac{d}{2}\); from \(R_h = \frac{b \cdot d}{b + 2d}\), \(\frac{d}{2} = \frac{d}{2}\) confirms this valid for maximum efficiency alongside condition P.

In conclusion, the correct set of conditions for maximum discharge are P and S. Hence, the correct answer is:

P and S.