Simply Simplify!!

$\dfrac {1}{ \sqrt{2} } \left( \dfrac {x+2} {x-2} + \dfrac {x + 2 \sqrt{2} } {x - 2 \sqrt{2} } \right) = \sqrt{k} \quad \Rightarrow \dfrac {x+2} {x-2} + \dfrac {x + 2 \sqrt{2} } {x - 2 \sqrt{2} } = \sqrt{2k}$

Substitute in the LHS of the above equation $x = \frac {4 \sqrt{2}} {\sqrt{2} +1}$

$\dfrac { \frac {4 \sqrt{2}} {\sqrt{2} +1} +2} { \frac {4 \sqrt{2}} {\sqrt{2} +1} -2} + \dfrac { \frac {4 \sqrt{2}} {\sqrt{2} +1} + 2 \sqrt{2} } {\frac {4 \sqrt{2}} {\sqrt{2} +1} - 2 \sqrt{2} }$

$= \dfrac { 4 \sqrt{2} + 2 \sqrt{2} + 2 } { 4 \sqrt{2} - 2 \sqrt{2} - 2} + \dfrac { 4 \sqrt{2} + 4 + 2 \sqrt{2} } { 4 \sqrt{2} - 4 - 2 \sqrt{2}}$

$= \dfrac { 6 \sqrt{2} + 2 } { 2 \sqrt{2} - 2} + \dfrac { 6 \sqrt{2} + 4 } { 2 \sqrt{2} - 4 }$

$= \dfrac { 3 \sqrt{2} + 1 } { \sqrt{2} - 1} + \dfrac { 3 \sqrt{2} + 2 } { \sqrt{2} - 2 }$

$= \dfrac { (3 \sqrt{2} + 1) ( \sqrt{2} - 2 ) + ( 3 \sqrt{2} + 2 ) ( \sqrt{2} - 1 ) } { (\sqrt{2} - 1) ( \sqrt{2} - 2 ) }$

$= \dfrac { 6 - 5 \sqrt{2} - 2 + 6 - \sqrt{2} - 2 } { 2 - 3 \sqrt{2} + 2 }$

$= \dfrac { 8 - 6 \sqrt{2} } { 4 - 3 \sqrt{2} } = 2$

$\Rightarrow \sqrt{2k} = 2 \quad \Rightarrow k = \boxed {2}$