Equation 11.27 in M.D. Schwartz book

I am having trouble understanding the steps taken in (11.27) equation in Quantum Field Theory and the Standard Model by M.D.Schwartz.

I don’t understand how to get the middle diagonal matrix in the second last statement of the equation.

Here is my attempt. I am writing
$$ begin{pmatrix} sqrt{p cdot sigma} xi_{s’} sqrt{p cdot bar sigma} xi_{s’} end{pmatrix} = begin{pmatrix} sqrt{p cdot sigma} quad 0 0 quad sqrt{p cdot bar sigma} end{pmatrix} begin{pmatrix}xi_{s’} xi_{s’} end{pmatrix} quad
and
quad begin{pmatrix} sqrt{p cdot sigma} xi_{s} sqrt{p cdot bar sigma} xi_{s} end{pmatrix}^dagger = begin{pmatrix}xi_{s} xi_{s} end{pmatrix}^dagger begin{pmatrix} sqrt{p cdot sigma} quad 0 0 quad sqrt{p cdot bar sigma} end{pmatrix}^dagger $$

and then proceeding as follows:

$$ quad begin{pmatrix} sqrt{p cdot sigma} xi_s sqrt{p cdot bar sigma} xi_s end{pmatrix}^dagger begin{pmatrix} 0 mathbb{I}_2 mathbb{I}_2 0 end{pmatrix} begin{pmatrix} sqrt{p cdot sigma} xi_{s’} sqrt{p cdot bar sigma} xi_{s’} end{pmatrix} quad quad (x) $$

$$ = begin{pmatrix}xi_{s} xi_{s} end{pmatrix}^dagger begin{pmatrix} sqrt{p cdot sigma} quad 0 0 quad sqrt{p cdot bar sigma} end{pmatrix}^dagger
begin{pmatrix} 0 mathbb{I}_2 mathbb{I}_2 0 end{pmatrix}
begin{pmatrix} sqrt{p cdot sigma} quad 0 0 quad sqrt{p cdot bar sigma} end{pmatrix} begin{pmatrix}xi_{s’} xi_{s’} end{pmatrix} quad quad quad (a)$$

$$ = begin{pmatrix}xi_{s} xi_{s} end{pmatrix}^dagger begin{pmatrix} sqrt{p cdot sigma} quad 0 0 quad sqrt{p cdot bar sigma} end{pmatrix}
begin{pmatrix} 0 mathbb{I}_2 mathbb{I}_2 0 end{pmatrix}
begin{pmatrix} sqrt{p cdot sigma} quad 0 0 quad sqrt{p cdot bar sigma} end{pmatrix} begin{pmatrix}xi_{s’} xi_{s’} end{pmatrix} quad quad quad (b)$$

$$ = begin{pmatrix}xi_{s} xi_{s} end{pmatrix}^dagger begin{pmatrix} sqrt{p cdot sigma} quad 0 0 quad sqrt{p cdot bar sigma} end{pmatrix}
begin{pmatrix} 0 quad sqrt{p cdot bar sigma} sqrt{p cdot sigma} quad 0 end{pmatrix} begin{pmatrix}xi_{s’} xi_{s’} end{pmatrix} quad quad quad (c)$$

$$ = begin{pmatrix}xi_{s} xi_{s} end{pmatrix}^dagger
begin{pmatrix} 0 quad sqrt{(p cdot sigma )( p cdot bar sigma)} sqrt{(p cdot bar sigma )( p cdot sigma)} quad 0 end{pmatrix} begin{pmatrix}xi_{s’} xi_{s’} end{pmatrix} quad quad quad (d)$$

Clearly, my middle matrix is very different than the correct* one.
Can anyone please point out or hint what am I doing wrong? Or how the author obtained the middle diagonal matrix?

*Update:
The "correct" expression ( i.e the one given in the book is ):

$$ = begin{pmatrix}xi_{s} xi_{s} end{pmatrix}^dagger
begin{pmatrix} sqrt{(p cdot sigma )( p cdot bar sigma)} quad 0 0 quad sqrt{(p cdot sigma )( p cdot bar sigma)} end{pmatrix} begin{pmatrix}xi_{s’} xi_{s’} end{pmatrix} quad quad quad (y)$$

To summarize, equation 11.27 says:
$$ (x) = (y) = 2mdelta_{ss’}$$

My main concern is how did the author go from (x) to (y)?