We have to find a constant value with the values and the equation given. Doing so, we have:
[tex]\begin{gathered} p(a)=\frac{?}{a} \\ a\cdot p(a)=\text{? (Multiplying by a on both sides of the equation)} \\ 14\cdot4=\text{? (Replacing the values)} \\ 56=\text{?(Multiplying)} \end{gathered}[/tex]The value of the numerator must be 56.