(b) As given, $({a^2} + {b^2} + {c^2}){p^2} – 2(ab + bc + cd)p$

$ + ({b^2} + {c^2} + {d^2}) \le 0$ ….. $(i)$

But $L.H.S.$

$ = ({a^2}{p^2} – 2abp + {b^2}) + ({b^2}{p^2} – 2bcp + {c^2})$

$ + ({c^2}{p^2} – 2cdp + {d^2})$ $ = {(ap – b)^2} + {(bp – c)^2} + {(cp – d)^2} \ge 0$ …..$(ii)$

Since the sum of squares of real numbers is non-negative.

Therefore from $(i)$ and $(ii)$

$ \Rightarrow $${(ap – b)^2} + {(bp – c)^2} + {(cp – d)^2} = 0$

$ \Rightarrow $$ap – b = 0 = bp – c = cp – d \Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$

$\therefore $$a,\;b,\;c,\;d$ are in $G.P.$