f interchanges $\alpha$ and $\beta$ but fixes $\gamma$ and $\delta$ 
$f=(\alpha \beta), \; g=(\beta \gamma), \; h=(\gamma \delta)$
Is there a way to solve this that avoids finding what the permutations in options are ?
 A: We observe that each of the candidate permutations is the conjugate of a transposition.  For instance, looking at permutation $A$ we have
$$
f \circ g \circ h \circ g \circ f = (f \circ g) \circ h \circ (f \circ g)^{-1}.
$$
(For the last equality, we are using the fact that a transposition is its own inverse as well as the shoe-sock theorem.)
It is a general fact that, given $n$, if we conjugate a transposition by any element of the symmetric group $S_n$, then the result is a transposition.  The upshot is that each of your candidate permutations is a transposition.  Thus to verify if one of the candidate permutations is the transposition $(\alpha \delta)$ you need only determine the image of one element.  This already saves a lot of work, but we can further save work by noticing that in each candidate permutation, 
it is always the case that $f$ appears only once (B and D) or $h$ appears only once (A and C).  
In permutations $A$ and $C$, $\delta$ does not move until $h$ is applied, so you only need to determine what happens to $\delta$ from the application of $h$ onward.  For instance, in permutation $A$, we have
$$
\delta \overset{h}{\mapsto} \gamma \overset{g}{\mapsto} {\beta} \overset{f}{\mapsto} \alpha,
$$
so permutation $A$ is indeed the transposition $(\alpha \delta)$.  
Similarly, in permutations $B$ and $D$, $\alpha$ does not move until you apply $f$, so you can trace what happens to $\alpha$ from the application of $f$ onward.
