The answer is yes. Suppose the contrary and rescale the picture so that $r=1$. We may assume that $P_1$ and $P_3$ are the endpoints of the graph. There must be points on the graph that are outside the circle - otherwise the curvature at $P_2$ is at least 1. WLOG assume that there are points below the circle. The lower half of the circle is the graph of a function $f_0$ defined on an interval of length 2. Let $q$ be a point where $f_0-f$ attains its maximum, then $f'(q)=f_0'(q)$. We may assume that $f'(q)=f_0'(q)\ge 0$, otherwise reflect the picture through the $y$-axis.
I claim that
$f(t)<f_0(t)$ for all $t\ge q$ such that both $f(t)$ and $f_0(t)$ are defined, contrary to the fact that $P_2$ lies on the circle. To prove this, consider arc-length parametrizations $\gamma(t)=(x(t),y(t))$ and $\gamma_0(t)=(x_0(t),y_0(t))$ of the two graphs, where $\gamma(0)=(q,f(q))$ and $\gamma_0(0)=(q,f_0(q))$. Then
where $0\le\alpha<\pi/2$. Since the curvature of $\gamma$ is less than 1, we have
\angle(\gamma'(t),\gamma'(0)) < t
for all $t>0$. Therefore $x'(t)>\cos(\alpha+t)$ and $y'(t)<\sin(\alpha+t)$ for
$0<t<\pi/2-\alpha$. And for $\gamma_0$ these inequalities turn to equalities. The integration yields that $x(t)\ge x_0(t)$ and
$y(t)< y_0(t)$ for all
$t\in[0,\pi/2-\alpha]$. Since $f_0$ is increasing after $q$, these inequalities imply that $\gamma(t)$ is below the half-circle (or has already left the domain where $f_0$ is defined).
Since $\gamma_0(\pi/2-\alpha)$ is the rightmost point of the circle, the inequality $x(t)\ge x_0(t)$ for $t=\pi/2-\alpha$ means that the $x$-coordinate of $\gamma(\pi/2-\alpha)$ is already outside the domain of $f_0$ and one does not need to care about larger values of $t$.