If the lines $$x = ay + b$$, $$z = cy + d$$ and $$x = a'z + b'$$, $$y = c'z + d'$$ are perpendicular, then:

We are given two straight lines in three-dimensional space.

The first line is described by the simultaneous equations

$$x = ay + b, \qquad z = cy + d.$$

We rewrite the coordinates of a general point on this line in vector form. Using the parameter $$y$$ itself as the free variable, we have

$$ (x,\;y,\;z)=\bigl(ay+b,\;y,\;cy+d\bigr). $$

Separating the constant part from the part multiplied by the parameter $$y$$ gives

$$ (x,\;y,\;z)=\underbrace{(b,\;0,\;d)}_{\text{a fixed point}} \;+\; y\;\underbrace{(a,\;1,\;c)}_{\text{direction vector}}. $$

Hence the direction vector of the first line is

$$ \vec v_1=(a,\;1,\;c). $$

Now we examine the second line, given by

$$x = a'z + b', \qquad y = c'z + d'.$$

This time we choose $$z$$ itself as the parameter. Then

$$ (x,\;y,\;z)=\bigl(a'z+b',\;c'z+d',\;z\bigr). $$

Again we separate the constant part and the part accompanied by the parameter $$z$$:

$$ (x,\;y,\;z)=\underbrace{(b',\;d',\;0)}_{\text{a fixed point}} \;+\; z\;\underbrace{(a',\;c',\;1)}_{\text{direction vector}}. $$

Therefore the direction vector of the second line is

$$ \vec v_2=(a',\;c',\;1). $$

For two lines in space to be perpendicular, their direction vectors must be perpendicular. The condition for perpendicularity of two vectors is that their dot product is zero. Explicitly, for vectors $$\vec u=(u_1,u_2,u_3)$$ and $$\vec v=(v_1,v_2,v_3)$$ we use the formula

$$ \vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3. $$

Applying this to $$\vec v_1=(a,1,c)$$ and $$\vec v_2=(a',c',1)$$, we compute

$$ \vec v_1\cdot\vec v_2 = a\;a' + 1\;c' + c\;1 = aa' + c' + c. $$

Setting this equal to zero for perpendicularity gives

$$ aa' + c + c' = 0. $$

This is exactly the relation stated in Option B.

Hence, the correct answer is Option B.