So I know the Boltzmann distribution is: $$ P\propto \exp \left(-E / k_BT \right) $$
where $E$ is energy, $k_B$ is the Boltzmann constant and $T$ is the temperature. However, when we replace $E$ for the kinetic energy $1/2 mv^2$ and we get $$ P \propto \exp \left(-mv^2 / 2k_BT \right) \, . $$
This becomes the probability for a particular velocity $\mathbf{v}$, and not for all particles with a set kinetic energy.
Why is this?
this becomes the probability for a particular velocity v, and not for all particles with a set kinetic energy.
Why is this?
No, I think you are misunderstanding the initial formulation of the Maxwell-Boltzmann distribution. Your first version with $P \propto e^{-E/kT}$ is a probability as a function of variable kinetic energy, $E$, for each particle and a single, scalar temperature, $T$, for the entire distribution.
The second version with $P \propto e^{-m \ v^{2}/2kT}$ is a probability as a function of variable velocity, v, for each particle and a single, scalar temperature, $T$, for the entire distribution.
Thus, in both cases one has $P$ as a function of some variable that is specific to each particle (i.e., $E$ or v) and a scalar (i.e., $T$) that is related to the full width at half max (FWHM) of the distribution.
Side Notes:
Because the formula depends on $v^2$, not $\vec{v}$, it's actually a probability for the particle to have a given speed. However, there's only one speed for each kinetic energy, so it's really the same as the probability for the particle to have the corresponding kinetic energy.
The probability for a particle to have a given velocity is vastly smaller, because there are so many velocities with a given speed (the particle can be moving in a vast number of possible directions).
