If $\mathrm{y}=4\mathrm{x}-5$  is a tangent to the curve ${\mathrm{y}}^2={\mathrm{px}}^3+\mathrm{q}$  at (2, 3) then

$\begin{array}{l}\mathrm{y}=4\mathrm{x}-5\\ \mathrm{m}=4\\ {\mathrm{y}}^2={\mathrm{px}}^3+\mathrm{q}\end{array}$
Diff.  w.r.t. ‘x’
$\begin{array}{l}2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{px}}^2\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3\mathrm{p}}{2\mathrm{y}}{\mathrm{x}}^2\\ \mathrm{m}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(2,3)}=\frac{3\mathrm{p}}{2\times 3}\times 4\\ =2\mathrm{p}\\ 2\mathrm{p}=4\Rightarrow \mathrm{p}=2\end{array}$
Equation of curve is 
${\mathrm{y}}^2=2{\mathrm{x}}^3+\mathrm{q}$
$\because \left(2,3\right)$ lie on the curve
$9=16+\mathrm{q}\Rightarrow \mathrm{q}=-7$