Step 1: Write the general Langmuir-Hinshelwood rate law.
$$\text{rate}=k\theta,\qquad \theta=\frac{Kp}{1+Kp}$$
This equation already shows the surface follows a Langmuir isotherm, not the empirical Freundlich isotherm proposed in option (A), so (A) is false right away.
Step 2: Take the two limiting cases of pressure.
High pressure ($Kp\gg1$): $1+Kp\approx Kp$, so $\theta\approx1$ and rate $\approx k$, independent of $p$. A rate independent of concentration is zero order, so option (B) holds.
Low pressure ($Kp\ll1$): $1+Kp\approx1$, so $\theta\approx Kp$ and rate $\approx kKp$, directly proportional to $p$. A rate directly proportional to concentration is first order, so option (C) holds.
Step 3: Read off option (D) directly.
The rate law rate $=k\theta$ already says the rate is proportional to the surface coverage $\theta$, which is exactly option (D); it is simply the defining relationship of the mechanism, so it is true.
Step 4: Physical picture behind the two limits.
At high pressure nearly every active site is filled, so adding more gas cannot speed the reaction further, giving the pressure-independent regime. At low pressure most sites are empty, so the number of adsorbed molecules grows directly with the gas supplied, giving the first order regime.
Final Answer:
Since the mechanism runs on a Langmuir isotherm, with zero order kinetics at high pressure, first order kinetics at low pressure, and the rate always tracking the surface coverage, the true statements are (B), (C), and (D).
\[ \boxed{\text{(B), (C), (D)}} \]