Answer
$\frac{1}{32}$ of its original value.
Work Step by Step
Step 1. Based on the given conditions, we have $p=kr^3/t^2$ where $k$ is a constant.
Step 2. If $r'=r/2, t'=2t$, we have $p'=k(r/2)^3/(2t)^2=\frac{1}{32}p$
Step 3. Thus, $p$ will be $\frac{1}{32}$ of its original value.