$$\begin{gathered} \because f^{\text{'}}\left(x^2\right)=\frac{1}{x} \\ \Rightarrow f^{\text{'}}\left(x\right)=\frac{1}{\sqrt{x}},x>0 \\ \Rightarrow f\left(x\right)=2\sqrt{x}+c \end{gathered}$$
      ( $c=$ integration constant)
 $$\begin{gathered} \because f\left(1\right)=1 \\ \Rightarrow c=-1 \end{gathered}$$
 $\therefore f\left(x\right)=2\sqrt{x}-1,x>0$
 And   $g^{\text{'}}({\sin }^{2}x-1)={\cos }^{2}x+p\forall x\in R$
 $$\begin{gathered} \Rightarrow g^{\text{'}}(-{\cos }^{2}x)={\cos }^{2}x+p \\ \Rightarrow g^{\text{'}}\left(x\right)=p-x,\forall x\in [-1,0] \\ \Rightarrow g\left(x\right)=px-\frac{x^2}{2}+k \end{gathered}$$
    (where $k=$ integration constant)
 $\because g(-1)=0\Rightarrow 0=-p-\frac{1}{2}+k\Rightarrow k=\frac{1}{2}+p$
 $$\begin{gathered} \therefore g\left(x\right)=px-\frac{x^2}{2}+\frac{1}{2}+p \\ \therefore h\left(x\right)=\{\begin{array}{cc}2\sqrt{x}-1,& x>0\\ px-\frac{x^2}{2}+\frac{1}{2}+p,& -1\le x\le 0\end{array} \end{gathered}$$
    At $x=0$ 
L.H.L = R.H.L =  $f\left(0\right)$
 $$\begin{gathered} \Rightarrow -1=\frac{1}{2}+p \\ \Rightarrow p=\frac{-3}{2} \end{gathered}$$
Hence  $2p=-3$
   Absolute value of  $2p$ is 3.