diagonals . . .

Join $BD$ and $CA$

Applying cosine rule on triangle $BAD$

$BD^2$ $=$ $AB^2$ $+$ $AD^2$ $+$ $2.AB.AD.cosA$ [ Equation 1 ]

Similarly, for triangle $ADC$

$AC^2$ $=$ $AD^2$ + $CD^2$ + $2.AD.CD. cosD$

Now note $AB=CD=5$ $AD=BC=8$ And angle $D$ = $180-A$

So $AC^2$ $=$ $AD^2$ + $CD^2$ + $2.AD.2.AD.CD. cosD$ becomes

$AC^2$ $=$ $AD^2$ + $AB^2$ + $2.AD.AB. cos(180-A)$ [ Equation 2]

Note $cos ( 180-A) = - cosA$

Adding equation $1$ and $2$

$BD^2 + AC^2 = 2 ( AB^2 + AD^2) +2AD.AB.cosA -2.AD.AB.cosA$

$BD^2+AC^2=2 (5^2 +8^2)$

$= 178$

NOTE: All the angles are in degrees

The sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its sides. In mathematical notation, $(AB)^{2} + (BC)^{2} + (CD)^{2} + (AD)^{2} = (BD)^{2} + (AC)^{2}$ Therefore, $(BD)^{2} + (AC)^{2} = 25 + 64 + 25 + 64 = \boxed{178}$