First of all, remember that dividing by a fraction is the same as multiplying by the inverse fraction:
\frac{p^2-11p+28}{2p} div \frac{p-7}{2p} = \frac{p^2-11p+28}{2p} \cdot \frac{2p}{p-7}
Cross simplify common terms:
\frac{p^2-11p+28}{cancel(2p)} \cdot \frac{cancel(2p)}{p-7} = \frac{p^2-11p+28}{p-7}
You can also observe that p^2-11p+28 is a quadratic equation. If we find the two roots p_1 and p_2, we may write p^2-11p+28 = (p-p_1)(p-p_2).
We can avoid the quadratic formula in this case: if the quadratic coefficient is 1 (which is our case), the equation can be read as
p^2-sx+p
where s is the sum of the solutions and p is their product. So, we're looking for two numbers which sum up to 11 and give 28 when multiplied. These numbers are clearly 4 and 7. So, you have
p^2-11p+28 = (p-4)(p-7)
The fraction becomes
\frac{p^2-11p+28}{p-7} = \frac{(p-4)cancel((p-7))}{cancel(p-7)} = p-4
