Answer :
Use the PMF for the given distribution:
[tex]P(X=x) = \begin{cases}\dbinom{15}x 0.8^x (1-0.8)^{15-x} & \text{if } x \in\{0, 1, 2, \ldots, 15\} \\ 0 & \text{otherwise}\end{cases}[/tex]
Then the probability that X = 14 is
[tex]P(X=14) = \dbinom{15}{14} 0.8^{14} 0.2^1 \approx \boxed{0.1319}[/tex]