First, we will find #MP#.
Because #MNOP# is a rectangle, we know that #bar(MP)# is parallel to #bar(ON)#, and thus to #bar(BC)#. This implies that #angleAMP = angleABC# and #angleAPM = angle ACB#, meaning #triangleAMP# is similar to #triangleABC#, and so is also isosceles.
As #AM = MB# and #AM+MB = s#, we know that #s = 2AM#, or #AM = s/2#. Because #triangleAMP# is isosceles, this also gives us #AP = s/2#. Using the Pythagorean theorem, then, we have #MP^2 = AM^2 + AP^2 = 2(s/2)^2 = s^2/2#, and so #MP = s/sqrt(2)#.
Next, we will find #MN#.
Because #MNOP# is a rectangle, we know #angleMNO=90^@#. Then, as #angleBNM# is its compliment, we also have #angleBNM = 90^@#.
As the non-right angles of an isosceles right triangle are #45^@#, we know #angleABC = 45^@#, implying #angleMBN = 45^@#. Thus #triangleBNM# is also an isosceles right triangle, and so #BN = NM#.
Applying the Pythagorean theorem again, we have #BM^2 = BN^2 + MN^2 = 2MN^2#. But, as #BM = s/2#, we can substitute that in and solve for #MN# to obtain #MN = s/(2sqrt(2))#
Now that we have the side lengths of the rectangle, we can easily find its perimeter #p# and area #A#.
#p = 2(s/sqrt(2)) + 2(s/(2sqrt(2))) = (2s)/sqrt(2)+s/sqrt(2) = 3/sqrt(2)s#
#A = (s/sqrt(2))(s/(2sqrt(2))) = s^2/4#
