If two numbers $p$ and $q$ are chosen randomly from the set $\{1,2,3,4\}$, one by one, with replacement, then the probability of getting $p^2>4 q$ is

Two numbers are selected with replacement from $S = \{1,2,3,4\}$, giving $4 \times 4 = 16$ total ordered pairs $(p,q)$.

The condition $p^2 \ge 4q$ is analyzed:

When $p=1$, $p^2=1$; no $q$ satisfies $1 \ge 4q$.

When $p=2$, $p^2=4$; the condition $4 \ge 4q$ holds only for $q=1$.

When $p=3$, $p^2=9$; $9 \ge 4q$ yields favorable $q \in \{1,2\}$.

When $p=4$, $p^2=16$; all $q \in \{1,2,3,4\}$ satisfy $16 \ge 4q$.

The number of favorable outcomes is $0 + 1 + 2 + 4 = 7$.

The probability is $\frac{7}{16}$, corresponding to option B.