jmc

We know that ${\left(\frac{13}{7}\right)}^2=\frac{169}{49}$. Then, since $3=\frac{147}{49}<\frac{169}{49}<\frac{196}{49}=4$, we conclude that $\lceil {\left(\frac{13}{7}\right)}^2\rceil =4$. Because $4+\frac{17}{4}=\frac{33}{4}$, which is a number between $8$ and $9$, $\lfloor \lceil {\left(\frac{13}{7}\right)}^2\rceil +\frac{17}{4}\rfloor =\boxed{8}$.