From the point (-1, -1) , two rays are sent making angles of 45°...

The given line is $x + y = 0$, which has a slope of $m_1 = -1$.

Let the slope of an incident ray be $m$. Since it makes an angle of $45^{\circ} with the line$x + y = 0$, we have:<br>\left| \dfrac{m - (-1)}{1 + m(-1)} \right| = \tan 45^{\circ} = 1$
$\Rightarrow \dfrac{m + 1}{1 - m} = \pm 1$

Taking $+1$: $m + 1 = 1 - m \Rightarrow 2m = 0 \Rightarrow m = 0$
Taking $-1$: $m + 1 = -1 + m \Rightarrow 1 = -1$ (which is not possible, implying $m \to \infty$).

The incident rays pass through $P(-1, -1)$. Thus, their equations are:
Ray 1: $y = -1$
Ray 2: $x = -1$

Next, we find the points of intersection of these incident rays with the mirror $x + 2y = 1$.
For Ray 1 ($y = -1$): $x + 2(-1) = 1 \Rightarrow x = 3$. The point of incidence is $A(3, -1)$.
For Ray 2 ($x = -1$): $-1 + 2y = 1 \Rightarrow 2y = 2 \Rightarrow y = 1$. The point of incidence is $B(-1, 1)$.

The reflected rays will appear to originate from the image of $P(-1, -1)$ in the mirror $x + 2y - 1 = 0$. Let the image be $P'(x', y')$.
$\dfrac{x' - (-1)}{1} = \dfrac{y' - (-1)}{2} = -2 \dfrac{1(-1) + 2(-1) - 1}{1^2 + 2^2}$
$\dfrac{x' + 1}{1} = \dfrac{y' + 1}{2} = -2 \left(\dfrac{-4}{5}\right) = \dfrac{8}{5}$
$\Rightarrow x' = \dfrac{8}{5} - 1 = \dfrac{3}{5}$ and $y' = \dfrac{16}{5} - 1 = \dfrac{11}{5}$
So, $P' = \left(\dfrac{3}{5}, \dfrac{11}{5}\right)$.

The reflected rays pass through $P'$ and their respective points of incidence.

Equation of Reflected Ray 1 (passing through $A and P'$):
Slope $m_1' = \dfrac{\dfrac{11}{5} - (-1)}{\dfrac{3}{5} - 3} = \dfrac{\dfrac{16}{5}}{-\dfrac{12}{5}} = -\dfrac{4}{3}$
$y - (-1) = -\dfrac{4}{3}(x - 3) \Rightarrow 3y + 3 = -4x + 12 \Rightarrow 4x + 3y = 9$
Comparing with $ax + by = 9$, we get $a = 4, b = 3$.

Equation of Reflected Ray 2 (passing through $B and P'$):
Slope $m_2' = \dfrac{\dfrac{11}{5} - 1}{\dfrac{3}{5} - (-1)} = \dfrac{\dfrac{6}{5}}{\dfrac{8}{5}} = \dfrac{3}{4}$
$y - 1 = \dfrac{3}{4}(x - (-1)) \Rightarrow 4y - 4 = 3x + 3 \Rightarrow 3x - 4y = -7 \Rightarrow -3x + 4y = 7$
Comparing with $cx + dy = 7$, we get $c = -3, d = 4$.

Finally, we calculate $ad + bc$:
$ad + bc = (4)(4) + (3)(-3) = 16 - 9 = 7$

Answer: $7$