Confusion in barometer experiment
Said by other words, you do not believe that $3 + 2 = 3 + 2$

There are 2 $\ce{Hg}$ columns, sharing the same pressure at the bottom.
Both have the same height, so the pressure difference is the same.
Yet you have doubts the pressures at their tops are the same.


Pressure at A = Pressure at B +pgh

It should be $p_\mathrm{A} = p_\mathrm{B} + \rho gh$ where $\rho$ is the density of mercury.

The vertical gradient of the hydrostatic pressure in liquid of the density $\rho$ at gravitational accelerationg $g$ is:
$$\mathrm{d}p = - g \cdot \rho \cdot \mathrm{d}h$$
As we can consider $\rho$ and $g$ as constants, the difference of the pressures is then :
$$p_C - p_A = - g \cdot \rho \cdot \int_{h_A}^{h_C} {\mathrm{d}h}$$
$$h_C=h_A \implies p_C=p_A$$

If pressure at the same altitude within the connected liquid of constant density was different, there would be the liquid flow until it is equal.
The pressures at the same altitude would be slightly different, if e.g. mercury at both columns had different temperature and density.

Feedback to comments:
In fact, it does not matter if it is mercury or water, the important thing is the density is constant across the liquid. What other than atmospheric pressure would you expect at C ( regardless of the liquid below ) ? Your scenario description is not exactly clear. Now at A, there is pressure by mercury column which is nothing but it's weight.
If the density along the thought path from A down and then up to C is constant, then pressure at A and C are the same, if the liquid is in the rest. Any pressure difference would cause motion of the liquid.

Let assume the liquid is mercury with density $\rho$.
Let assume the atmospheric pressure at C is $p_0$.
Let assign to C the relative height $h=0$.
Then pressure anywhere below C ( with negative $h$ ) is
$$ p = p_0 - \rho \cdot g \cdot h \gt p_0$$
The same is anywhere below A. At A is the same pressure as at C, i.e. $p_0$.
Between A and B, at height $h$, is the pressure:
$$ p = p_0 - \rho \cdot g \cdot h \lt p_0,$$
but this time $h$ is positive.
B is then at the height
$$ h_\mathrm{B}= \frac {p_0}{\rho \cdot g}$$
with $p=0$
( if we neglect very low vapour pressure of mercury.)