Answer: Option D
$$\eqalign{
& {\text{Give,}} \cr
& p + \frac{1}{{p + 2}} = 1 \cr
& {\text{Adding 2 both sides}} \cr
& p + 2 + \frac{1}{{p + 2}} = 1 + 2 = 3 \cr
& {\text{Let}}\left( {p + 2} \right) = a\,\,\&\,\, \frac{1}{{p + 2}} = b \cr
& a + b = 3 \cr
& {\text{Cubbing both sides}} \cr
& {\left( {a + b} \right)^3} = {3^3} \cr
& {a^3} + {b^3} + 3ab\left( {a + b} \right) = 27 \cr
& {a^3} + {b^3} + 3 \times ab \times 3 = 27 \cr
& {a^3} + {b^3} + 9ab = 27......\left( {\text{i}} \right) \cr
& {\text{Now,}} \cr
& a \times b = \left( {p + 2} \right) \times \frac{1}{{\left( {p + 2} \right)}} \cr
& a \times b = 1........\left( {{\text{ii}}} \right) \cr
& {\text{Put the a & b value in equation}}\left( {\text{i}} \right) \cr
& {a^3} + {b^3} + 9 \times 1 = 27 \cr
& {a^3} + {b^3} = 27 - 9 = 18 \cr
& \therefore {\left( {p + 2} \right)^3} + \frac{1}{{{{\left( {p + 2} \right)}^3}}} - 3 \cr
& = {a^3} + {b^3} - 3 \cr
& = 18 - 3 \cr
& = 15 \cr} $$