$$k \frac{d}{dx}[A(x)\frac{dT(x)}{dx}] - hP(x)[T(x) - T] = 0 $$
What I had in mind was: $$x = rcosϴ, r = \frac{x}{cosϴ} , \frac{dr}{dx} = \frac{1}{cosϴ} $$ $$\frac{dA(x)}{dx} = \frac{dA(r)}{dx}\frac{dr}{dx} = \frac{1}{cosϴ}\frac{dA(r)}{dr}$$ $$\frac{dT(x)}{dx} = \frac{dT(r)}{dx}\frac{dr}{dx} = \frac{1}{cosϴ}\frac{dT(r)}{dr}$$ $$\frac{d²T(x)}{dx²} = \frac{d}{dr}[\frac{dT(r)}{dr}\frac{1}{cosϴ}]\frac{dr}{dx} = \frac{1}{cosϴ}\frac{d²T(r)}{dr²}$$ $$\frac{k}{cos²ϴ}\frac{d}{dr}[A(r)\frac{dT(r)}{dr}] - hP(r)[T(r) - T] = 0 $$ but I'm getting the sensation that this is wrong.
Polar coordinates are two-dimensional; your problem has just one spatial variable, so that's clearly not the way to go. However, the your ODE is second order, so it can be written as a two-dimensional system of first order ODEs: \begin{align} A(x)\frac{\text{d} T}{\text{d} x} &= S(x),\\ k \frac{\text{d} S}{\text{d} x} &= h P(x) \left(T(x)-T\right). \end{align} You can now try to convert $(T,S)$ to polar coordinates, i.e. write \begin{align} T(x) &= r(x)\,\cos[ \theta(x)],\\ S(x) &= r(x)\,\sin[ \theta(x)], \end{align} and substitute this in the system to get equations for $\frac{\text{d} r}{\text{d} x}$ and $\frac{\text{d} \theta}{\text{d} x}$. However, as long as $P(x)$ is unknown, this is of limited value.