I have attached both the problem and approach to the solution. But I am confused how to proceed. Kindly help in providing the simplified version of the solution.
PS- I tried pasting the problem and solution from a word file, but I don't understand why the editor is not accepting my text. That's why I have attached the images.
I'm not at all sure that the following remarks are in any real sense a simplification of the argument given in our OP Sanket Garud's question and answer, but perhaps I can expand upon the essential points and thus provide some clarification.
Given a unit speed curve $\gamma(s)$ in $\Bbb R^3$, where $s$ is the arc-length, its unit tangent vector $T(s)$ is defined via
$T(s) = \dot \gamma(s); \tag 1$
the unit normal vector $N(s)$ and curvature $\kappa(s)$ are defined by the first Frenet-Serret equation
$\dot T = \kappa N, \; \vert N \vert = 1, \; \kappa(s) > 0; \tag 2$
we also have the unit binormal vector
$B(s) = T(s) \times N(s); \tag 3$
with the hypothesis $\tau(s) = 0$, the third Frenet-Serret equation yields
$\dot B(s) = -\tau(s) N(s) = -(0)N(s) = 0; \tag 4$
that is, $B(s)$ is constant along $\gamma(s)$; we thus adopt the shorthand
$B = B(s); \tag{4.5}$
in accord with the definition (3) we have
$B \cdot N(s) = T(s) \times N(s) \cdot N(s) = 0 \tag 5$
and
$B \cdot T(s) = T(s) \times N(s) \cdot N(s) = 0. \tag 6$
Now everything which has been written in the above is either a well-known and accepted fact of elementary differential geometry or has been covered by our OP in his question/answer. We next deviate from the presented work of Sanket Garud and observe that (1) implies
$\gamma(s) - \gamma(s_0) = \displaystyle \int_{s_0}^s T(u) \; du, \tag 7$
whence
$B \cdot (\gamma(s) - \gamma(s_0)) = B \cdot \displaystyle \int_{s_0}^s T(u) \; du = \int_{s_0}^s B \cdot T(u) \; du = \int_{s_0}^s 0 \; du = 0, \tag 8$
which shows that $\gamma(s) - \gamma(s_0)$ lies in a plane normal to $B$; thus
$\gamma(s) = (\gamma(s) - \gamma(s_0)) + \gamma(s_0) \tag{8.5}$
lies in the plane normal to $B$ and containing $\gamma(s_0)$; furthermore, in accord with (5) and (6), both $T$ and $N$ lie in, are tangent to, this plane at every $\gamma(s)$; thus the vector
$C(s) = \gamma(s) + \dfrac{1}{\kappa}N(s) \tag 9$
lies in this plane as well, since $N(s)$ does; we have
$\dot C(s) = \dot \gamma(s) + \dfrac{1}{\kappa} \dot N(s) = \dot \gamma(s) + \dfrac{1}{\kappa} \dot N(s)$ $= T(s) + \dfrac{1}{\kappa} (-\kappa T(s) + \tau B(s)) = T(s) + \dfrac{1}{\kappa}(-\kappa T(s)) = T(s) - T(s) = 0; \tag{10}$
thus,
$C(s) = C \tag{11}$
is constant, and (9) becomes
$\gamma(s) = C - \dfrac{1}{\kappa}N(s); \tag{12}$
thus,
$\vert \gamma(s) - C \vert = \left \vert \dfrac{1}{\kappa} N(s) \right \vert = \dfrac{1}{\kappa} \vert N(s) \vert = \dfrac{1}{\kappa}, \tag{13}$
which shows that $\gamma(s)$ lies in the circle of radius $1/\kappa$ centered at $C$ in the plane normal to $B$ containing $\gamma(0)$. We note that if $\gamma(s)$ is not restricted to lie in this plane, then (13) still binds but now tells us that $\gamma(s)$ lies in the sphere with the same radius and center. Since both the circle and sphere described by (13) are of the same radius $1/\kappa$, the circle is in fact a great circle in this sphere.