Pressure is caused by containment, and not weight per se, but weight or gravity causes containment, so, in a way, weight can be said to generate pressure.
When weight initiates pressure though, unlike other containments, there is a gradient in pressure by height.
And, once generated, pressure can remain there as long as containment is there.
This can be proven by imagining enclosing the earth with a shell $30$ km in height from earth's surface, and then turning off gravity.
The atmosphere will retain its total average pressure, but the gradient will be gone.
That's how pressure in a small bottle remains equal to $1$ atm even if we seal it from all sides, and take it to outer space.
The adjustment you talked about is more in terms of number of molecules than their velocity. The number of molecules both inside and outside are equal and that's why the pressure is equal.
How Does Gravity Generate Pressure?
The following discussion assumes that there is enough gravity so that air molecules don't have escape velocity at normal temperatures.
In the presence of gravity, momentum of every air molecule increases as it moves downward. This means that every molecule, say, in layer A, would feel more force when it was hit from above than when it was hit from below. This would gradually move the molecules of layer A downward until a stage is reached when this effect is cancelled by there being more molecules in the lower layer.
This means that the density of lower layers is more in the presence of gravity, which, in turn, means that the pressure becomes more as we descend. This is how gravity (or weight) generates pressure with a gradient.
(An interesting thing to note is that this model of molecules bouncing in a layer generates a downward force that is equal to the combined weight of the molecules in that layer plus of those above it!
Otherwise it was possible to pay less for a can of talcum powder by shaking it vigorously before taking it to the counter and then showing it is lighter than what the label says.)
This is somewhat similar to how pressure will be generated if in space, we start pushing in the piston of a cylinder (filled with air and closed at other end).
In both the cases above, once we have generated the pressure, if we
- turn off gravity (while maintaining the containment using a shell), or
- we stop moving the piston (but hold it there),
the generated pressure will remain.
Why does the pressure need to be equalized?
If we mix water dyed with different colors, why do the colors mix? It's the same thing happening here. Because there are more randomly moving molecules in a region than its vicinity, there will be an equalization. This is because of entropy and the second law of thermodynamics.
Can $P=h\rho g$ be shown equal to $P=1/3\rho v^2$?
$P=h\rho g$ is not true for a compressible gas with varying density (such as atmosphere).
The formula that is actually true and that results in $P=h\rho g$ for constant $\rho$ is $\frac{dP}{dh} = -\rho g$.
Both these formulae ($\frac{dP}{dh} = -\rho g$ and $P=1/3\rho v^2$) are true separately, and, interestingly $P=1/3\rho v^2$ can be shown to yield $\frac{dP}{dh} = -\rho g$ in the presence of g (see below).
So no, these formula cannot be shown equal.
Bonus
The kinetic formula $P=1/3\rho v^2$ yields the hydrostatic formula $\frac{dP}{dh} = \rho g$. Use $\rho$ from eqn. (7) here assuming T (and v) = constant. We get: $$
\begin{align}
\frac{d\rho}{dh} &= -\rho \frac{gm}{kT} \\
\implies \frac{dP}{dh} &= \frac{1}{3}v^2\frac{d\rho}{dh} \\
&= \frac{1}{3}v^2 \times (-\rho \frac{gm}{kT}) \\
&= - \frac{1}{3} \times \frac{3kT}{m} \times \rho \frac{gm}{kT} \\
&= -\rho g
\end{align}
$$
