\[p(b) = \left( b + \frac{5}{b} \right)\left( 5b + \frac{1}{b} \right);\]
\[p\left( \frac{1}{b} \right) = \left( \frac{1}{b} + 5b \right)\left( \frac{5}{b} + b \right)\]
\[\frac{p(b)}{p\left( \frac{1}{b} \right)} = \frac{\left( b + \frac{5}{b} \right)\left( 5b + \frac{1}{b} \right)}{\left( \frac{1}{b} + 5b \right)\left( \frac{5}{b} + b \right)} = 1.\]