In a common emitter configuration with suitable bias, it is given than $$R_L$$ is the load resistance and $$R_{BE}$$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by: [$$\beta$$ is current gain, $$I_B$$, $$I_C$$, $$I_E$$ are respectively base, collector and emitter currents]

We begin with the small-signal definition of voltage gain in a common-emitter (CE) amplifier. By definition, the voltage gain is

$$A_v=\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}.$$

In the CE configuration the output voltage is taken across the collector load $$R_L$$, while the input voltage is the small-signal base-emitter voltage $$\Delta V_{BE}$$. Hence we can write

$$A_v=\frac{\Delta V_C}{\Delta V_{BE}}.$$

Now, on the output side, a small change in collector current $$\Delta I_C$$ flowing through the load produces a change in the collector voltage given by Ohm’s law:

$$\Delta V_C=-\Delta I_C\,R_L.$$

(The negative sign merely indicates a phase reversal; for gain magnitude we can drop it.) On the input side, the change in base-emitter voltage is related to the change in base current $$\Delta I_B$$ through the small-signal dynamic resistance $$R_{BE}$$:

$$\Delta V_{BE}=\Delta I_B\,R_{BE}.$$

Substituting these two expressions into the definition of $$A_v$$ we obtain

$$A_v=\frac{-\Delta I_C\,R_L}{\Delta I_B\,R_{BE}}.$$

In a CE transistor, small-signal collector current and base current are connected by the current gain $$\beta$$, so we have the important relation

$$\Delta I_C=\beta\,\Delta I_B.$$

Replacing $$\Delta I_C$$ in the previous expression gives

$$A_v=\frac{-\beta\,\Delta I_B\,R_L}{\Delta I_B\,R_{BE}}.$$

The factor $$\Delta I_B$$ cancels, and ignoring the negative sign for magnitude we get

$$A_v=\beta\,\frac{R_L}{R_{BE}}.$$

Thus the magnitude of the voltage gain is $$\beta R_L/R_{BE}$$.

Next we turn to current gain. In the CE configuration the relevant small-signal current gain is defined as

$$A_i=\frac{\Delta I_C}{\Delta I_B}.$$

But by definition of $$\beta$$ in a transistor, $$\beta = \Delta I_C/\Delta I_B$$. Therefore

$$A_i=\beta=\frac{\Delta I_C}{\Delta I_B}.$$

Finally, the power gain is simply the product of voltage gain and current gain because power is the product of voltage and current. Therefore

$$A_p=A_v\,A_i=\left(\beta\,\frac{R_L}{R_{BE}}\right)\times\beta=\beta^2\,\frac{R_L}{R_{BE}}.$$

Collecting the three results together, we have

$$\text{Voltage gain}= \beta\frac{R_L}{R_{BE}},\qquad \text{Current gain}= \frac{\Delta I_C}{\Delta I_B},\qquad \text{Power gain}= \beta^2\frac{R_L}{R_{BE}}.$$

Comparing these expressions with the choices given, we see that they match exactly with Option D.

Hence, the correct answer is Option D.