The correct answer is B. < P.
The power dissipated in a wire is given by the equation $P = I^2R$, where $I$ is the current, $R$ is the resistance, and $P$ is the power. The resistance of a wire is given by the equation $R = \rho \frac{l}{A}$, where $\rho$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the area of the cross-section of the wire.
The resistivity of copper is $\rho_c = 1.68 \times 10^{-8} \Omega \cdot m$, and the resistivity of aluminium is $\rho_a = 2.65 \times 10^{-8} \Omega \cdot m$. Therefore, the resistance of an aluminium wire is greater than the resistance of a copper wire of the same length and area of cross-section.
Since the power dissipated is inversely proportional to the resistance, the power dissipated in an aluminium wire is less than the power dissipated in a copper wire of the same length and area of cross-section.