In a triangle ABC with usual notations if $\angle A=30^{\circ}$, then the value of $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

Given expression: $E = \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$

Combine terms over common denominators:

$E = \left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right) = \frac{(a+b+c)(b+c-a)}{bc}$

Recognize the numerator as a difference of squares where $X = b+c$ and $Y = a$:

$E = \frac{(b+c)^{2} - a^{2}}{bc}$

Expanding gives:

$E = \frac{b^{2} + c^{2} + 2bc - a^{2}}{bc}$

Using the Law of Cosines, $a^{2} = b^{2} + c^{2} - 2bc \cos A$, so $b^{2} + c^{2} - a^{2} = 2bc \cos A$. Substituting:

$E = \frac{2bc \cos A + 2bc}{bc} = 2(\cos A + 1)$

Given $A = 30^{\circ}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, then:

$E = 2\left(\frac{\sqrt{3}}{2} + 1\right) = \sqrt{3} + 2$

This matches option C.

Final answer: $\boxed{\text{C}}$