Transfering moment through internal hinge

My professor says, the internal hinge can not transfer moment:

But can we calculate $\Sigma M_A=0$ for whole beam ( without cutting it or something) this way?:

$\Sigma M_A=0 ===\Rightarrow M_A+(D_y \times 4m ) + 3KN.M =0$

with this equation we literally transferred the 3KN.M moment to point A while my professor says the internal hinge can not transfer moment.

Something is wrong, and I don't know what it is.

"Hinges don't transfer moment" is just a way of saying that hinges do not resist bending, and therefore suffer no bending moment.

You can basically "translate" that term as "Bending moment at hinges is always zero."

Just note that in more complex structures such as frames there may be hinges which only affect some members but not others on the same node. In that case, there may still be moment on the node, but the hinged beam will have zero moment on that node, such as below (see the hinge on the central node only affecting the left-hand span):

The internal bending moment at the hinge must be zero, but the hinge can still transmit a force into the AB cantilever. If you consider a plain cantilever with point force at the end, you have not applied a moment, but there is a moment reaction force at the base - that’s what is happening here.

The moment 3KN creates two equal and opposite forces at supports B and D.

Each is equal to $3KN/(Length- of-BD).$ With Dy pushing up and By pushing down.

D-y is countered by the D upward reaction and cancels.

The correct moment at A is $$M_a = (\frac {3KN}{length-of-BD} \times AB)$$ which is clockwise. $$M_a = force_B \times (length AB)$$

And your teacher is right, no moment is transferred, but shear has been transferred.