Let the directrix of the parabola $P: y^2 = 8x$, cut $x$-axis at the point $A$. Let $B(\alpha, \beta)$, $\alpha > 1$, be a point on $P$ such that the slope of $AB$ is $3/5$. If $BC$ is a focal chord of $P$, then six times the area of $\triangle ABC$ is :

For the parabola $P: y^2 = 8x$, we have $4a = 8 \Rightarrow a = 2$.

The equation of the directrix is $x = -a \Rightarrow x = -2$. Since the directrix cuts the $x$-axis at $A$, the coordinates of $A$ are $(-2, 0)$.

Let the coordinates of point $B$ on the parabola be $(2t_1^2, 4t_1)$.
The slope of $AB$ is given as $\dfrac{3}{5}$.

$\dfrac{4t_1 - 0}{2t_1^2 - (-2)} = \dfrac{3}{5}$

$\dfrac{2t_1}{t_1^2 + 1} = \dfrac{3}{5}$

$10t_1 = 3t_1^2 + 3 \Rightarrow 3t_1^2 - 10t_1 + 3 = 0$

$(3t_1 - 1)(t_1 - 3) = 0 \Rightarrow t_1 = \dfrac{1}{3}$ or $t_1 = 3$.

For $t_1 = \dfrac{1}{3}$, $\alpha = 2\left(\dfrac{1}{3}\right)^2 = \dfrac{2}{9}$, which is rejected since $\alpha > 1$.

For $t_1 = 3$, $\alpha = 2(3)^2 = 18 > 1$. Thus, $B$ is $(18, 12)$.

Since $BC$ is a focal chord, the parameter for $C$ is $t_2 = -\dfrac{1}{t_1} = -\dfrac{1}{3}$.

The coordinates of $C$ are $\left(2\left(-\dfrac{1}{3}\right)^2, 4\left(-\dfrac{1}{3}\right)\right) = \left(\dfrac{2}{9}, -\dfrac{4}{3}\right)$.

The area of $\triangle ABC$ is given by:
$\Delta = \dfrac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$

$\Delta = \dfrac{1}{2} \left| -2\left(12 - \left(-\dfrac{4}{3}\right)\right) + 18\left(-\dfrac{4}{3} - 0\right) + \dfrac{2}{9}(0 - 12) \right|$

$\Delta = \dfrac{1}{2} \left| -2\left(\dfrac{40}{3}\right) - 24 - \dfrac{24}{9} \right|$

$\Delta = \dfrac{1}{2} \left| -\dfrac{80}{3} - 24 - \dfrac{8}{3} \right| = \dfrac{1}{2} \left| -\dfrac{88}{3} - \dfrac{72}{3} \right| = \dfrac{1}{2} \left| -\dfrac{160}{3} \right| = \dfrac{80}{3}$

Six times the area of $\triangle ABC = 6 \times \dfrac{80}{3} = 160$.

Answer: $160$