It's not so complex...

As given in the question, $\left| \beta \right| =1$

But, $\beta \bar { \beta } ={ \left| \beta \right| }^{ 2 }\\ \beta \bar { \beta } =1$

Now in the given equation, put $\beta =\frac { 1 }{ \bar { \beta } }$

So, $\left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| =\left| \frac { \frac { 1 }{ \bar { \beta } } -\alpha }{ 1-\bar { \alpha } \frac { 1 }{ \bar { \beta } } } \right| \\ \\ \left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| =\left| \frac { 1-\alpha \bar { \beta } }{ \bar { \beta } -\bar { \alpha } } \right| \\ \\ \left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| \left| \frac { \bar { \beta } -\bar { \alpha } }{ 1-\alpha \bar { \beta } } \right| =1\\$

Now, clearly from the last expression you can see that the two terms are conjugates of each other. So, their product can be given by,

${ \left| \left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| \right| }^{ 2 }=1$

Now, since this is the square of a positive number, so,

$\left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| =1$

CHEERS!!:)

$|\beta|=1$ or, $\beta \bar{\beta}=1$ .

Now, $|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}|$ $= |\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha}\beta}|$ $= |\frac{\beta-\alpha}{\beta (\bar{\beta}-\bar{\alpha})}|$ $= \frac{1}{|\beta|}\times \frac{|\beta-\alpha|}{| \bar{\beta}-\bar{\alpha}|}$ $=\boxed{1}$