Hint: \({p}=\frac{\Delta {I}}{{c}}\)

Step 1: Find the radiation pressure.
Radiation pressure \((p)\) is the force exerted by an electromagnetic wave on
the unit area of the surface, i.e., the rate of change of momentum per unit area of the surface.
The momentum per unit time per unit area is given by; 
\(\Rightarrow \frac{\text { Intensity }}{\text { Speed of wave }}=\frac{{I}}{{c}}\)
Then, radiation pressure is; \(p= \frac{\Delta I}{c}\)
Also, the momentum of the incident wave per unit time per unit area = \(\frac{{I}}{{c}}\)

Step 2: Find the radiation pressure for the absorbed wave.
When the wave is fully absorbed by the surface, the momentum of the reflected
wave per unit time per unit area \(=0\)
Then, radiation pressure \((p)= \) change in momentum per unit time per unit area 
\(\Rightarrow \frac{\Delta {I}}{{c}}=\frac{{I}}{{c}}-0=\frac{{I}}{{c}}\)

Step 3: Find the radiation pressure for the reflected wave.
When the wave is totally reflected, then the momentum of the reflected wave per
unit time per unit area = \(-\frac{I}{ c}\)
The radiation pressure will be; \(\Rightarrow \frac{I}{c}-\left(-\frac{I}{c}\right)=\frac{2 I}{c}\)
It is evident that  \( p\) lies between \(\frac{I}{c}\) and \(\frac{2I}{c}\) for a real surface, as the wave is partially reflected
by a real surface. Therefore, statements a, c, and d are valid under the given conditions.
Hence, option (3) is the correct answer.