jmc

By Vieta's formulas, the sum of the roots of $p(x)$ is 0, so the third root is $t=-r-s.$ Also, $$a=rs+rt+st.$$The sum of the roots of $q(x)$ is also 0, so the third root is $-(r+4)-(s-3)=-r-s-1=t-1.$ Also, $$a=(r+4)(s-3)+(r+4)(t-1)+(s-3)(t-1).$$Hence, $$rs+rt+st=(r+4)(s-3)+(r+4)(t-1)+(s-3)(t-1).$$This simplifies to $t=4r-3s+13.$

Also, $b=-rst$ and $$b+240=-(r+4)(s-3)(t-1).$$Hence, $$-rst+240=(r+4)(s-3)(t-1).$$Substituting $t=4r-3s+13,$ we get $$-rs(4r-3s+13)+240=-(r+4)(s-3)(4r-3s+12).$$This simplifies to $$r^2-2rs+s^2+7r-7s-8=0.$$Then $(r-s{)}^2+7(r-s)-8=0,$ which factors as $$(r-s-1)(r-s+8)=0.$$Thus, $r-s=1$ or $r-s=-8.$

If $r-s=1,$ then $s=r-1,$ and $$t=4t-3s+13=r+16.$$But $r+s+t=0,$ so $r+(r-1)+(r+16)=0,$ which leads to $r=-5.$ Then $s=-6$ and $t=11,$ and $b=-rst=-330.$

If $r-s=-8,$ then $s=r+8,$ and $$t=4t-3s+13=r-11.$$But $r+s+t=0,$ so $r+(r+8)+(r-11)=0,$ which leads to $r=1.$ Then $s=9$ and $t=-10,$ and $b=-rst=90.$

Thus, the possible values of $b$ are $\boxed{-330,90}.$