The links provided by bubba indeed answer the question, but I'll follow through with the simple-minded approach I outlined in the comment.
Choose coordinates $x,y,z$ so that $p$ is the origin and the tangent vector to $C$ at the origin is $(1,0,0)$. This implies that the curve can be parametrized by $\gamma(t)=(t,bt^2,ct^2)+o(t^2)$ where $o(t^2)$ does not affect the curvature at $p$. Indeed,
$$\kappa^2 = 4(b^2+c^2)\tag1$$
so we only need to find $b$ and $c$.
Let $f=0$ and $g=0$ be equations of the given surfaces. Observe that $f\circ \gamma$ and $g\circ \gamma $ are identically zero. Taking the first derivative in $t$, we find (unsurprisingly) that $f_x=0$ and $g_x=0$ at the origin.
Now look at the second-order Taylor expansion of $f\circ \gamma$ and $g\circ \gamma $: equating the coefficient of $t^2$ to zero, we find
$$\begin{cases} \frac12 f_{xx}+bf_y+cf_z &=0 \\ \\ \frac12 g_{xx}+bg_y+cg_z &=0 \end{cases}\tag2$$
Here and in what follows all derivatives are evaluated at $p$. By Cramer's rule,
$$b = -\frac12 \frac{f_{xx} g_z-g_{xx}f_z}{f_yg_z-f_zg_y} \quad
c = -\frac12 \frac{g_{xx} f_y-f_{xx}g_y}{f_yg_z-f_zg_y} \tag3$$
Plug (3) into (1) to find
$$\kappa^2 = \frac{f_{xx}^2|\nabla g|^2+g_{xx}^2 |\nabla f|^2 -2f_{xx}g_{xx}\nabla f\cdot \nabla g}{(f_yg_z-f_zg_y)^2}\tag4$$
It remains to interpret the derivatives in (4) geometrically. Namely, $$
\begin{align}
f_{xx}&=\lambda_1|\nabla f| \\ \\ g_{xx}& =\lambda_2|\nabla g| \\ \\
\nabla f\cdot \nabla g &= |\nabla f||\nabla g|\cos\theta \\ \\
|f_yg_z-f_zg_y| &= |\nabla f||\nabla g|\sin\theta \end{align}\tag5$$
which after some magic cancellation of gradients transforms (4) into
$$\kappa^2 = \frac{\lambda_1^2+\lambda_2^2 -2\lambda_1\lambda_2 \cos\theta }{\sin^2\theta}\tag6$$
