I don't understand the case 4 of red-black tree deletion
but both case {3,4} will not change the pointer , anyway so how case 4 work?


Let us check case 4. Although the parent pointer of node $x$ is not changed, i.e., the parent node of $x$ is still $B$, the color of $B$ is set to black and node $D$ is lifted to become $B$'s parent with $B$'s original color. 

So, going from the graph on the left hand side of case 4 on the image in the question to the graph on the right hand side, we have added that "extra" black node that was supposed to attached to $x$, which is, in your words, an extra black node $B$ is passed for the paths going through $\alpha,\beta$. 

Furthermore, for that graph on the right hand side, we can check that all simple paths from the node to descendant leaves that do not pass through $A$ contain the same number of black nodes as before. Since all properties of black-red tree have been restored, case 4 ends the while loop, which ends RB-DELETE-FIXUP$(T, x)$, which ends RB-DELETE$(T, z)$.