$$\begin{gathered} \mathrm{Suppose} \mathrm{X}~\mathrm{B}\left(\mathrm{n}, \mathrm{p}\right) \mathrm{and} \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(\mathrm{X}=5\right). \\ \mathrm{If} \mathrm{p}>\frac{1}{2} \mathrm{then} \end{gathered}$$

Given X follows Binomial distribution.

Given that $\mathrm{P}(X=3)=\mathrm{P}(X=5)$

$$\begin{gathered} {}^{\mathrm{n}}{\mathrm{c}}_3 {\mathrm{p}}^3 {\mathrm{q}}^{\mathrm{n}-3}={}^{\mathrm{n}}{\mathrm{c}}_5 9^5 {\mathrm{q}}^{\mathrm{n}-5} \\ \Rightarrow {}^{\mathrm{n}}{\mathrm{c}}_3 p^3 {(1-p)}^{\mathrm{n}-3}={}^{\mathrm{n}}{\mathrm{c}}_5 p^5 {(1-p)}^{\mathrm{n}-5} \\ \Rightarrow \frac{(\mathrm{n}-3) (\mathrm{n}-4)}{20}=\frac{{(1-p)}^2}{{\mathrm{P}}^2}<1 \left(\because p>\frac{1}{2}\right) \\ \Rightarrow \frac{(\mathrm{n}-3) (\mathrm{n}-4)}{20}<1 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\mathrm{n}}^2-7\mathrm{n}+12<20 \\ \Rightarrow {\mathrm{n}}^2-7\mathrm{n}-8<0 \\ \Rightarrow {\mathrm{n}}^2-8\mathrm{n}+\mathrm{n}-8<0 \\ \Rightarrow \mathrm{n}(\mathrm{n}-8)+1(\mathrm{n}-8)<0 \\ \Rightarrow (\mathrm{n}-8) (\mathrm{n}+1)<0 \\ \Rightarrow -1<\mathrm{n}<8 \end{gathered}$$

By verifing options 

i.e.  $5\le \mathrm{n}\le 7$