CLRS red black tree deletion - red-black-tree

Question

01 void RedBlackTree::deleteNode(Node *z)
02 {
03 Node *y = z;
04 Color yOriginalColor = y->color;
05 Node *x;
06 if (z->left == sentinel)
07 {
08 x = z->right;
09 transplant(z, z->right);
10 }
11 else if (z->right == sentinel)
12 {
13 x = z->left;
14 transplant(z, z->left);
15 }
16 else
17 {
18 y = minimum(z->right);
19 yOriginalColor = y->color;
20 x = y->right;
21 if (y->parent == z)
22 {
23 x->parent = y; // WHY??
24 }
25 else
26 {
27 transplant(y, y->right);
28 y->right = z->right;
29 y->right->parent = y;
30 }
31 transplant(z, y);
32 y->left = z->left;
33 y->left->parent = y;
34 y->color = z->color;
35 }
36
37 delete z;
38
39 if (yOriginalColor == Color::BLACK)
40 {
41 deleteFixUp(x);
42 }
43 }
In line 20, we assign y->right to x. In line 23, we assign y to x->parent.
Isn't the assignment in line 23 redundant since the parent of x is y already?
In the book it says
When y's original parent is z, however, we do not want x.p to point to y's original parent, since we are removing that node from the tree. Because node y will move up to take z's position in the tree, setting x.p to y in line 13 causes x.p to point to the original position of y's parent, even if x = T.nil.
I don't understand the explanation from the book, why would x.p points to z. When we move node y from its original position to z's position, shouldn't y's child x remain unaffected, still attached to y?
I come across 2 posts asking similar queations as I do, however, I'm still confused after reading through the answers.
red black tree pseudocode redundancy
issue with the red black delete algorithm mentioned in introduction to algorithm
After these 2 questions, I get one more question. I don't understand the assignment of y to x.p in line 23 has anything to do with x being the sentinel node (e.g. T.nil) or not.