This is a check on the second to last line of page 218.  
A proof that replacing  (((Subscript[u, k]))^((r)))(z)  with   (((Subscript[u, n - k]))^((r)))(z)  replaces (Subscript[P, 5])  with  (((-1))^((2  n - 1) n))(Subscript[P, 5])  is obtained by noting that:  to place the first n rows in reverse order and then place the last n rows in reverse order requires an even number of permutations; then, to place the resulting 2n columns in reverse order requires 
       (2n-1) + (2n-2) + ... + 1 = (2n-1)n  permutations.

$Version

n1 =  1

A1 = Det[(({{a0, a1}, {b0, b1}}))]

B1= Det[(({{a1, a0}, {b1, b0}}))]

k1 = (-1)^(2*n1 - 1)*n1

A1 - (-1)^k1*B1

n2 = 2

A2 = Det[(({{a0, a1, a2, 0}, {0, a0, a1, a2}, {b0, b1, b2, 0}, {0, b0, b1, b2}}))]

B2 = Det[(({{a2, a1, a0, 0}, {0, a2, a1, a0}, {b2, b1, b0, 0}, {0, b2, b1, b0}}))]

k2 = (2  n2 - 1)*n2

A2 - (-1)^k2*B2

n3 = 3

A3=Det[(({{a0, a1, a2, a3, 0, 0}, {0, a0, a1, a2, a3, 0}, {0, 0, a0, a1, a2, a3}, {b0, b1, b2, b3, 0, 0}, {0, b0, b1, b2, b3, 0}, {0, 0, b0, b1, b2, b3}}))]

B3 = Det[(({{a3, a2, a1, a0, 0, 0}, {0, a3, a2, a1, a0, 0}, {0, 0, a3, a2, a1, a0}, {b3, b2, b1, b0, 0, 0}, {0, b3, b2, b1, b0, 0}, {0, 0, b3, b2, b1, b0}}))]

k3 = (2*n3 - 1) n3

A3 - (-1)^k3*B3

n4 = 4

A4=Det[(({{a0, a1, a2, a3, a4, 0, 0, 0}, {0, a0, a1, a2, a3, a4, 0, 0}, {0, 0, a0, a1, a2, a3, a4, 0}, {0, 0, 0, a0, a1, a2, a3, a4}, {b0, b1, b2, b3, b4, 0, 0, 0}, {0, b0, b1, b2, b3, b4, 0, 0}, {0, 0, b0, b1, b2, b3, b4, 0}, {0, 0, 0, b0, b1, b2, b3, b4}}))];

B4=Det[(({{a4, a3, a2, a1, a0, 0, 0, 0}, {0, a4, a3, a2, a1, a0, 0, 0}, {0, 0, a4, a3, a2, a1, a0, 0}, {0, 0, 0, a4, a3, a2, a1, a0}, {b4, b3, b2, b1, b0, 0, 0, 0}, {0, b4, b3, b2, b1, b0, 0, 0}, {0, 0, b4, b3, b2, b1, b0, 0}, {0, 0, 0, b4, b3, b2, b1, b0}}))];

k4 = (2  n4 - 1) n4

A4 - (-1)^k4*B4  // Expand

Date[]

n5 = 5

Date[]

A5 = Det[(({{a0, a1, a2, a3, a4, a5, 0, 0, 0, 0}, {0, a0, a1, a2, a3, a4, a5, 0, 0, 0}, {0, 0, a0, a1, a2, a3, a4, a5, 0, 0}, {0, 0, 0, a0, a1, a2, a3, a4, a5, 0}, {0, 0, 0, 0, a0, a1, a2, a3, a4, a5}, {b0, b1, b2, b3, b4, b5, 0, 0, 0, 0}, {0, b0, b1, b2, b3, b4, b5, 0, 0, 0}, {0, 0, b0, b1, b2, b3, b4, b5, 0, 0}, {0, 0, 0, b0, b1, b2, b3, b4, b5, 0}, {0, 0, 0, 0, b0, b1, b2, b3, b4, b5}}))];

Date[]

B5=Det[(({{a5, a4, a3, a2, a1, a0, 0, 0, 0, 0}, {0, a5, a4, a3, a2, a1, a0, 0, 0, 0}, {0, 0, a5, a4, a3, a2, a1, a0, 0, 0}, {0, 0, 0, a5, a4, a3, a2, a1, a0, 0}, {0, 0, 0, 0, a5, a4, a3, a2, a1, a0}, {b5, b4, b3, b2, b1, b0, 0, 0, 0, 0}, {0, b5, b4, b3, b2, b1, b0, 0, 0, 0}, {0, 0, b5, b4, b3, b2, b1, b0, 0, 0}, {0, 0, 0, b5, b4, b3, b2, b1, b0, 0}, {0, 0, 0, 0, b5, b4, b3, b2, b1, b0}}))];

Date[]

k5 = (2  n5 - 1) n5

A5 - (-1)^k5*B5  // Expand

?Resultant

Date[]

C5 = 
        Resultant[a0*X^5 + a1*X^4 + a2*X^3 + a3*X^2 + a4*X + a5, 
          b0*X^5 + b1*X^4 + b2*X^3 + b3*X^2 + b4*X + b5, X];

Date[]

D5 = 
        Resultant[a5*X^5 + a4*X^4 + a3*X^3 + a2*X^2 + a1*X + a0, 
          b5*X^5 + b4*X^4 + b3*X^3 + b2*X^2 + b1*X + b0, X];

Date[]

A5 - C5   // Expand

B5 - D5   // Expand

Date[]

At the next step, the computation appears to be too elaborate and remains unevaluated.

n6  =  6

A6=Det[(({{a0, a1, a2, a3, a4, a5, a6, 0, 0, 0, 0, 0}, {0, a0, a1, a2, a3, a4, a5, a6, 0, 0, 0, 0}, {0, 0, a0, a1, a2, a3, a4, a5, a6, 0, 0, 0}, {0, 0, 0, a0, a1, a2, a3, a4, a5, a6, 0, 0}, {0, 0, 0, 0, a0, a1, a2, a3, a4, a5, a6, 0}, {0, 0, 0, 0, 0, a0, a1, a2, a3, a4, a5, a6}, {b0, b1, b2, b3, b4, b5, b6, 0, 0, 0, 0, 0}, {0, b0, b1, b2, b3, b4, b5, b6, 0, 0, 0, 0}, {0, 0, b0, b1, b2, b3, b4, b5, b6, 0, 0, 0}, {0, 0, 0, b0, b1, b2, b3, b4, b5, b6, 0, 0}, {0, 0, 0, 0, b0, b1, b2, b3, b4, b5, b6, 0}, {0, 0, 0, 0, 0, b0, b1, b2, b3, b4, b5, b6}}))];

Date[]

B6=Det[(({{a6, a5, a4, a3, a2, a1, a0, 0, 0, 0, 0, 0}, {0, a6, a5, a4, a3, a2, a1, a0, 0, 0, 0, 0}, {0, 0, a6, a5, a4, a3, a2, a1, a0, 0, 0, 0}, {0, 0, 0, a6, a5, a4, a3, a2, a1, a0, 0, 0}, {0, 0, 0, 0, a6, a5, a4, a3, a2, a1, a0, 0}, {0, 0, 0, 0, 0, a6, a5, a4, a3, a2, a1, a0}, {b6, b5, b4, b3, b2, b1, b0, 0, 0, 0, 0, 0}, {0, b6, b5, b4, b3, b2, b1, b0, 0, 0, 0, 0}, {0, 0, b6, b5, b4, b3, b2, b1, b0, 0, 0, 0}, {0, 0, 0, b6, b5, b4, b3, b2, b1, b0, 0, 0}, {0, 0, 0, 0, b6, b5, b4, b3, b2, b1, b0, 0}, {0, 0, 0, 0, 0, b6, b5, b4, b3, b2, b1, b0}}))];

Date[]

k6  =  (2  n6 - 1) n6

A6 - (-1)^k6*B6   // Expand

Date[]