If  $x^{3p}$ occurs in the expansion of ${\left(x+\frac{{\displaystyle 1}}{{\displaystyle x^3}}\right)}^{4n}$, then its coefficients is 

${\left(x+\frac{{\displaystyle 1}}{{\displaystyle x^3}}\right)}^{4n}$

$T_{r+1}={}^{4n}{\text{C}}_r{x}^{4n-r}{\left(\frac{{\displaystyle 1}}{{\displaystyle x^3}}\right)}^r$

$={}^{4n}{\text{C}}_r\frac{x^{4n-r}}{x^{3r}}$

$={}^{4n}{\text{C}}_r{x}^{4n-4r}$

$\text{for\hspace{0.17em}\hspace{0.17em}}{x}^{3p};4n-4r=3p$

$4r=4n-3p$

$r=\frac{4n-3p}{4}$

$\therefore x^{3p}\text{coefficient}={}^{4n}{\text{C}}_r$

$={}^{4n}{\text{C}}_{\frac{4n-3p}{4}}$

$=\frac{\left(4n\right)!}{\left(4n-\frac{4n-3p}{4}\right)!\left(\frac{4n-3p}{4}\right)!}$

$=\frac{\left(4n\right)!}{\left(\frac{12n+3p}{4}\right)!\left(\frac{4n-3p}{4}\right)!}$

$=\frac{\left(4n\right)!}{\left[\left(\frac{3}{4}\right)(4n+p)]![\frac{1}{4}(4n-3p)\right]!}$