jmc

Squaring the equation $p+q+r+s=8,$ we get $$p^2+q^2+r^2+s^2+2(pq+pr+ps+qr+qs+rs)=64.$$Hence, $p^2+q^2+r^2+s^2=64-2\cdot 12=40.$

By Cauchy-Schwarz, $$(1^2+1^2+1^2)(p^2+q^2+r^2)\ge (p+q+r{)}^2.$$Then $3(40-s^2)\ge (8-s{)}^2.$ Expanding, we get $120-3s^2\ge 64-16s+s^2,$ so $4s^2-16s-56\le 0.$ Dividing by 4, we get $s^2-4s-14\le 0.$ By the quadratic formula, the roots of the corresponding equation $x^2-4x-14=0$ are $$x=2\pm 3\sqrt{2},$$so $s\le 2+3\sqrt{2}.$

Equality occurs when $p=q=r=2-\sqrt{2},$ so the maximum value of $s$ is $\boxed{2+3\sqrt{2}}.$