The truth table for $(\sim p\lor q)$ is as follows:
$p$ |
$q$ |
$\sim p$ |
$(\sim p\lor q)$ |
T |
T |
F |
T |
T |
F |
F |
F |
F |
T |
T |
T |
F |
F |
T |
T |
from the above, it is clearly visible that it is $(p\to q)$ truth table.
$(p\to q)$ is a conditional statement. the truth table is shown below:
$p$ |
$q$ |
$(p\to q)$ |
T |
T |
T |
T |
F |
F |
F |
T |
T |
F |
F |
T |
Another approach:
$\because (p\lor q) \equiv (\sim p\to q)$
using above property we can rewrite the given expression:
$(\sim p\lor q) \ = [\sim(\sim p)\to q]$
$(\sim p\lor q) \equiv (p\to q)$
So option ($A)$ is correct.