$\because (p\lor q)=(\sim p\to q)$
$\therefore (\sim q\lor p)=(\sim(\sim q)\to p)\equiv (q\to p)$
we can also verify this with the truth table method:
$p$ |
$q$ |
$\sim q$ |
$(\sim q\lor p)$ |
$(p\to q)$ |
$(q\to p)$ |
$p\leftrightarrow q$ |
$p\lor q$ |
T |
T |
F |
T |
T |
T |
T |
T |
T |
F |
T |
T |
F |
T |
F |
T |
F |
T |
F |
F |
T |
F |
F |
T |
F |
F |
T |
T |
F |
F |
T |
F |
From above $(\sim q\lor p) \equiv (q\to p)$
So Option (B) is correct.