The eccentricity of the ellipse $\dfrac{(x-1)^2}{2} + \left(y + \dfrac{3}{4}\right)^2 = \dfrac{1}{16}$ is

$\dfrac{(x-1)^2}{2} + \left(y + \dfrac{3}{4}\right)^2 = \dfrac{1}{16}$

$\Rightarrow 8(x-1)^2 + 16\left(y +\dfrac{3}{4}\right)^2 = 1$

$\Rightarrow \dfrac{(x-1)^2}{\dfrac{1}{8}} + \dfrac{\left(y + \dfrac{3}{4}\right)^2}{\dfrac{1}{16}} = 1$

$\therefore a^2 = \dfrac{1}{8}$ and $b^2 = \dfrac{1}{16}$

$\Rightarrow a = \dfrac{1}{2\sqrt{2}}$ and $b = \dfrac{1}{4}$

$\therefore e = \sqrt{1 - \dfrac{b^2}{a^2}}$ $[\because a > b]$

$= \sqrt{1-\dfrac{\dfrac{1}{16}}{\dfrac{1}{8}}} = \sqrt{1 - \dfrac{1}{2}} = \dfrac{1}{\sqrt{2}}$