There are two questions to this problem:
1) Prove that the curve $\alpha(s)$ is a straight line.
2) Does the conclusion hold if $\alpha$ is not regular, ie. $\alpha'(s)=0$ for some $s\in I$.
I proved the first part:
Since the lines pass through a fixed point $p$ we can write the condition:
$$\alpha(s) +\lambda(s)t(s)=d$$
Taking the derivative on both sides we get:
$$\alpha'(s)+\lambda'(s)t(s)+\lambda(s)t'(s)=0$$
$$(1+\lambda'(s))t(s)+\lambda(s)k(s)n(s)=0$$
Since the vectors $t$ and $n$ are perpendicular we have $1+\lambda'(s)=0$ and $\lambda(s)k(s)=0$. Here it follows that $\lambda'(s)=-1$ and $k(s)=0 \implies t'(s)=0$. So $\lambda(s)=-s+c$ and $t(s)=t_0$. Hence we can rewrite the condition:
$$\alpha(s)=\underbrace{a_0}_{d-a}+st_0$$
which is a line.
What I am having trouble with is the 2nd question. My guess is that it should not be true but I don't see where the condition that the curve $\alpha(s)$ is regular has been used in the proof above. Clearly if $t=0$ at some point we can have a problem with the results that $\lambda'(s)=-1$ and $k(s)=0$ but these should be isolated points and not interfere with the general idea.
Best Answer
Think about pieces-wise linear curves like the graph of $|x|$. all tangents intersect at the origin. Any curve whose tangent line intersect at a point will have a trace which is a union of line segments with a common end point.