With usual notation, in triangle $\mathrm{ABC}, \mathrm{m} \angle \mathrm{A}=30^{\circ}$, then the value of $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$ is equal to

Let $E = \left(1 + \frac{a}{c} + \frac{b}{c}\right)\left(1 + \frac{c}{b} - \frac{a}{b}\right)$.

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

Applying the Law of Sines with circumradius $R$ yields $a = 2R\sin A$, $b = 2R\sin B$, $c = 2R\sin C$.

Substituting into the expression and simplifying:
$E = \frac{(\sin A + \sin B + \sin C)(\sin B + \sin C - \sin A)}{\sin B \sin C}$.

Applying the difference of squares formula:
$E = \frac{(\sin B + \sin C)^2 - \sin^2 A}{\sin B \sin C} = \frac{\sin^2 B + \sin^2 C + 2\sin B \sin C - \sin^2 A}{\sin B \sin C}$.

Separating terms: $E = \frac{\sin^2 B + \sin^2 C - \sin^2 A}{\sin B \sin C} + 2$.

From the Law of Cosines converted via the Law of Sines, $\sin^2 B + \sin^2 C - \sin^2 A = 2\sin B \sin C \cos A$.

Substituting gives $E = 2\cos A + 2$.

With $\angle A = 30^\circ$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$:
$E = 2\left(1 + \frac{\sqrt{3}}{2}\right) = 2 + \sqrt{3}$.

This matches option B.