⚡ AZUCATION SHORTCUT:
In a parallelogram, $P + R = Q + S$.
$S = P + R - Q = (-3+9-1, -2+1-(-5)) = (5, 4)$.
Line $SQ$ passes through $(1, -5)$ and $(5, 4)$.
Find $x$-intercept where $y=0$: $\frac{0 - (-5)}{x - 1} = \frac{4 - (-5)}{5 - 1} \implies \frac{5}{x-1} = \frac{9}{4} \implies 20 = 9x - 9 \implies x = \mathbf{29/9}$.
(Wait, checking calculation: $20+9=29$, so $29/9$ is Option B. Let's correct the key logic).

Step 1: Find Vertex $S$
In a parallelogram $PQRS$, the diagonals $PR$ and $QS$ bisect each other. Therefore, the midpoint of $PR$ is the same as the midpoint of $QS$.
Midpoint of $PR = \left(\frac{-3+9}{2}, \frac{-2+1}{2}\right) = (3, -0.5)$.
Let $S = (x_s, y_s)$. Midpoint of $QS = \left(\frac{1+x_s}{2}, \frac{-5+y_s}{2}\right)$.
Equating them: $1+x_s = 6 \implies x_s = 5$ and $-5+y_s = -1 \implies y_s = 4$. So, $S = (5, 4)$.

Step 2: Equation of Diagonal $SQ$
The diagonal $SQ$ connects $Q(1, -5)$ and $S(5, 4)$.
Slope ($m$) of $SQ = \frac{4 - (-5)}{5 - 1} = \frac{9}{4}$.
Equation: $y - 4 = \frac{9}{4}(x - 5) \implies 4y - 16 = 9x - 45 \implies 9x - 4y = 29$.

Step 3: Intersection with $x$-axis
The diagonal intersects the $x$-axis at $(a, 0)$. Substitute $y=0$ in the equation:
$9(a) - 4(0) = 29 \implies 9a = 29 \implies a = \mathbf{\frac{29}{9}}$.

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