The vertices of the triangle ABC are A (0,0), B(3, 0) and C(3, 4) ,...

$AC = 5 = 2ae$
$⇒ 2ae = 5 \ldots \left(i\right)$
Also $AB + BC = 2a$
$⇒ 2a = 7$ $⇒ a = \dfrac{7}{2}$
$∴ e = \dfrac{5}{7}$

$⇒ 1 - \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{25}{49} ⇒ b = \sqrt{6}$
Length of the latus rectum, $\dfrac{2{b}^{2}}{a} = \dfrac{2 \times 6}{\dfrac{7}{2}}$ units

$⇒ p = \dfrac{7}{2} = 3 .5$