Checking the validity of formula $\rm P = mv^{2}$.

STEP 1: Find the units of the physical quantities on the left-hand side of the equation.

$\rm P = \frac{Work}{Time} = \frac{kgm^{2}s^{-2}}{s} = kgm^{2}s^{-3}$

STEP 2: Find the units of the physical quantities on the right-hand side of the equation,

$\rm mv^{2} = kg \cdot \left ( \frac{m}{s} \right ) ^{2} = kgm^{2} s^{-2}$

From steps 1 and 2, we observe that the units of the terms in the given equation are different. Hence, this alternative formula for Power is invalid.

Checking the validity of formula $\rm P = \frac{mv}{A}$.

STEP 1: Find the units of the physical quantities on the left-hand side of the equation.

$\rm P = \frac{Force}{Area} = \frac{kgms^{-2}}{m^{2}} = kgm^{-1}s^{-2}$

STEP 2: Find the units of the physical quantities on the right-hand side of the equation,

$\rm \frac{mv}{A} = \frac{kg \cdot ms^{-1}}{m^{2}} = kg m^{-1} s^{-1}$

From steps 1 and 2, we observe that the units of the terms in the given equation are different. Hence, this alternative formula for Pressure is also invalid.