First, we can factorize p, then we have
[tex]\frac{1}{2}p-\frac{3}{8}p=(\frac{1}{2}-\frac{3}{8})p[/tex]The fraction in parenthesis is equal to
[tex]\begin{gathered} \frac{1}{2}-\frac{3}{8}=\frac{1\cdot8-3\cdot2}{2\cdot8} \\ \frac{1}{2}-\frac{3}{8}=\frac{8-6}{16} \\ \frac{1}{2}-\frac{3}{8}=\frac{2}{16} \\ \frac{1}{2}-\frac{3}{8}=\frac{1}{8} \end{gathered}[/tex]Hence, the answer is
[tex]\frac{1}{2}p-\frac{3}{8}p=\frac{1}{8}p[/tex]