(c) Given that ${a^2} + {b^2} = 1$, therefore
$\frac{{1 + b + ia}}{{1 + b – ia}} = \frac{{(1 + b + ia)(1 + b + ia)}}{{(1 + b – ia)(1 + b + ia)}}$
$ = \frac{{{{(1 + b)}^2} – {a^2} + 2ia(1 + b)}}{{1 + {b^2} + 2b + {a^2}}}$$ = \frac{{(1 – {a^2}) + 2b + {b^2} + 2ia(1 + b)}}{{2(1 + b)}}$
$ = \frac{{2{b^2} + 2b + 2ia(1 + b)}}{{2\,(1 + b)}} = b + ia$
Trick : Put $a = 0,b = 1$, $\frac{{1 + b + ia}}{{1 + b – ia}} = \frac{{1 + 1 + 0}}{{1 + 1 – 0}} = 1$
But options $(a)$ and $ (c)$  give $1.$
So again put $a = 1,b = 0,\frac{{1 + b + ia}}{{1 + b – ia}} = \frac{{1 + i}}{{1 – i}} = i$.
Which gives $ (c)$ only.