The difference between the larger root and the smaller root of $x^2 - px + (p^2 - 1)/4 = 0$ is: 
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ p\qquad\textbf{(E)}\ p+1$
Call the two roots $r$ and $s$, with $r \ge s$. 
By Vieta's formulas, $p=r+s$ and $(p^2-1)/4=rs.$
(Multiplying both sides of the second equation by 4 gives $p^2-1=4rs$.)
The value we need to find, then, is $r-s$.
Since $p=r+s$, $p^2=r^2+2rs+s^2$.
Subtracting $p^2-1=4rs$ from both sides gives $1=r^2-2rs+s^2$.
Taking square roots, $r-s=1 \Rightarrow \boxed{\textbf{(B)}}$.
(Another solution is to use the quadratic formula and see that the
roots are $\frac{p\pm 1}2$, and their difference is 1.)