If $f\left(x\right)=\left\{\begin{array}{ll}\frac{\sqrt{1+\sqrt{5+x}}-a}{\left(x-4\right)}& 0\le x<4\\ b& x\ge 4\end{array}\right.$ is continuous at $x=4$,
then value of  $\frac{1}{ab}$ is equal to

$\because f\left(x\right)$ is continuous at $x=4$
$\therefore \mathrm{}\mathrm{}\underset{x\to 4^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=\underset{x\to 4^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=f\left(4\right)$
$\underset{x\to 4^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sqrt{1+\sqrt{5+x}}-a}{x-4}=b$
for numerator to be zero $\Rightarrow a=2$
now after rationalization,
$\underset{x\to 4^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(\sqrt{1+\sqrt{5+x}}-2\right)\left(\sqrt{1+\sqrt{5+x}}+2\right)}{\left(x-4\right)\left(\sqrt{1+\sqrt{5+x}}+2\right)}=b$
$\underset{x\to 4^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1+\sqrt{5+x}-4}{\left(x-4\right)4}=b$
again on rationalization,

$\underset{x\to 4^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(5+x\right)-9}{\left(x-4\right)\times 4\left(\sqrt{5+x}+3\right)}=b$
$\frac{1}{4\times 6}=b\Rightarrow b=\frac{1}{24}$
$\frac{1}{ab}=\frac{24}{2}=12$