A function $f:R\to R$ is defined by $f\left(x\right)=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$.  Let the number of integral values of $\alpha$ for which $f$ is onto be $P$. Then the sum of digits of $P$ is

Since $f:R\to R$ is an onto mapping.
$\therefore$Range of f = R
$\Rightarrow \frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$ assumes all real value of x.
Let $y=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$
Then, y assumes all real values for real values of x. 
$\begin{array}{r}\Rightarrow \alpha y+6xy-8x^{2}y=\alpha x^2+6x-8\mathrm{\forall }y\in R\\ \Rightarrow x^2(\alpha +8y)+6x(1-y)-(8+\alpha y)=0\mathrm{\forall }y\in R\end{array}$
we know, above equation assumes all real values
$\begin{array}{l}\Rightarrow D\ge 0\\ \text{ So, }36(1-y{)}^2+4(\alpha +8y\left)\right(8+\alpha y)\ge 0\\ \Rightarrow 4\left[9\left(1-2y+y^2\right)+\left(8\alpha +{\alpha }^{2}y+64y+8\alpha y^2\right)\right]\ge 0\\ \Rightarrow \left[9-18y+9y^2+8\alpha +{\alpha }^{2}y+64y+8\alpha y^2\right]\ge 0\\ \Rightarrow \left[y^2(8\alpha +9)+y\left({\alpha }^2+46\right)+(8\alpha +9)\right]\ge 0\end{array}$
we know, if $\alpha x^2+bx+c>0\mathrm{\forall }x\text{ then }\mathrm{a}>0\text{ and }\mathrm{D}<0$
$\begin{array}{l}\text{ So, }{\left({\alpha }^2+46\right)}^2-4(8\alpha +9)(8\alpha +9)\le 0\text{ and }(8\alpha +9)>0\\ \Rightarrow {\left({\alpha }^2+46\right)}^2-\left[2\right(8\alpha +9\left)\right(8\alpha +9){]}^2\le 0\text{ and }\alpha >-9/8\\ \Rightarrow \left({\alpha }^2+46-16\alpha -18\right)\left({\alpha }^2+46+16\alpha +18\right)\le 0\text{ and }\alpha >-9/8\\ \left({\alpha }^2-16\alpha +28\right)\left({\alpha }^2+16\alpha +64\right)\le 0\text{ and }\alpha >-9/8\\ \Rightarrow (\alpha -14)(\alpha -2)(\alpha +8{)}^2\le 0\text{ and }\alpha >-9/8\\ \alpha \in [2,14]\cup \{-8\}\text{ and }\alpha >-9/8\end{array}$
Thus, $\alpha \in [2,14]$
Hence, f is onto when $\alpha \in [2,14]$