Answer
$$\frac{{dw}}{{dp}} = \frac{{{2^p}}}{{3\left( {{2^p} - 1} \right)}}$$
Work Step by Step
$$\eqalign{
& w = {\log _8}\left( {{2^p} - 1} \right) \cr
& {\text{differentiate with respect to }}p \cr
& \frac{{dw}}{{dp}} = \frac{d}{{dp}}\left[ {{{\log }_8}\left( {{2^p} - 1} \right)} \right] \cr
& {\text{use derivative}}{\log _a}x,\,\,\,\,\frac{d}{{dx}}\left[ {{{\log }_a}g\left( x \right)} \right] = \frac{1}{{\ln a}}\left( {\frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right).{\text{ set }}x = p,{\text{ }}a = 8 \cr
& \frac{{dw}}{{dp}} = \frac{1}{{\ln 8}}\left( {\frac{{\left( {{2^p} - 1} \right)'}}{{{2^p} - 1}}} \right) \cr
& \frac{{dw}}{{dp}} = \frac{1}{{\ln 8}}\left( {\frac{{{2^p}\ln 2}}{{{2^p} - 1}}} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dw}}{{dp}} = \frac{{\ln 2}}{{3\ln 2}}\left( {\frac{{{2^p}}}{{{2^p} - 1}}} \right) \cr
& \frac{{dw}}{{dp}} = \frac{{{2^p}}}{{3\left( {{2^p} - 1} \right)}} \cr} $$