Answer
$k^2-\dfrac{10}{7}kp+\dfrac{25}{49}p^2$
Work Step by Step
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $
\left( k-\dfrac{5}{7}p \right)^2
,$ is equivalent to
\begin{array}{l}\require{cancel}
(k)^2+2(k)\left(-\dfrac{5}{7}p\right)+\left(-\dfrac{5}{7}p\right)^2
\\\\=
k^2-\dfrac{10}{7}kp+\dfrac{25}{49}p^2
.\end{array}
