Let a focus of the ellipse E: (x^2)/(a^2) + (y^2)/(b^2) = 1 be S(4,...

Given the focus of the ellipse $S(4, 0)$, we have $ae = 4$.

Since the eccentricity $e = \dfrac{4}{5}$, we get $a \left(\dfrac{4}{5}\right) = 4 \Rightarrow a = 5$.

Using the relation $b^2 = a^2(1 - e^2)$, we find:

$b^2 = 25 \left(1 - \dfrac{16}{25}\right) = 9$

The equation of the ellipse is $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.

Since the point $P(3, \alpha)$ lies on the ellipse, substituting $x = 3$ gives:

$\dfrac{9}{25} + \dfrac{\alpha^2}{9} = 1$

$\dfrac{\alpha^2}{9} = 1 - \dfrac{9}{25} = \dfrac{16}{25}$

$\alpha^2 = \dfrac{144}{25} \Rightarrow |\alpha| = \dfrac{12}{5}$

The coordinates of the vertices of $\triangle POS$ are $O(0, 0)$, $S(4, 0)$, and $P\left(3, \pm \dfrac{12}{5}\right)$.

The area of $\triangle POS$ is $\dfrac{1}{2} \times \text{base} \times \text{height}$.

Taking $OS$ as the base, the length is $4$, and the height is the absolute value of the y-coordinate of $P$, which is $\dfrac{12}{5}$.

Area $= \dfrac{1}{2} \times 4 \times \dfrac{12}{5} = \dfrac{24}{5}$

Answer: $24/5$