Consider the parabola $P : y^2 = 4kx$ and the ellipse $E : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Let the line segment joining the points of intersection of $P$ and $E$, be their latus rectums. If the eccentricity of $E$ is $e$, then $e^2 + 2\sqrt{2}$ is equal to _____.

The latus rectum of the parabola $P: y^2 = 4kx$ is the line segment $x = k$. The endpoints of this latus rectum are $(k, 2k)$ and $(k, -2k)$.

The latus rectum of the ellipse $E: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is the line segment $x = ae$ (taking the one in the positive $x$-axis). The endpoints of this latus rectum are $\left(ae, \dfrac{b^2}{a}\right)$ and $\left(ae, -\dfrac{b^2}{a}\right)$.

Since the line segment joining the points of intersection of $P$ and $E$ is the latus rectum for both curves, their endpoints must coincide. Comparing the coordinates, we get:
$k = ae$
$2k = \dfrac{b^2}{a}$

Substituting $k = ae$ into the second equation:
$2ae = \dfrac{b^2}{a}$
$2a^2e = b^2$

For an ellipse, the relation between $a, b,$ and $e$ is $b^2 = a^2(1 - e^2)$. Substituting this into the equation above:
$2a^2e = a^2(1 - e^2)$

Since $a \neq 0$, we can divide by $a^2$:
$2e = 1 - e^2$
$e^2 + 2e - 1 = 0$

Solving for $e$ using the quadratic formula:
$e = \dfrac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \dfrac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$

Since the eccentricity of an ellipse must satisfy $0 < e < 1$, we take the positive root:
$e = \sqrt{2} - 1$

We need to find the value of $e^2 + 2\sqrt{2}$. First, calculate $e^2$:
$e^2 = (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$

Therefore:
$e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$

Answer: $3$