A particle of mass m is connected to three springs A , B , and C...

Let the mass be at the origin $(0,0)$.

The directions of the springs are:
Spring A: along the positive x-axis ($\hat{i}$)
Spring B: along the positive y-axis ($\hat{j}$)
Spring C: at $45^\circ$ to the negative x-axis, along $-\dfrac{1}{\sqrt{2}}\hat{i} - \dfrac{1}{\sqrt{2}}\hat{j}$

When the particle is displaced by a small distance $x$ against spring C, the displacement vector is:
$\vec{s} = x\left(-\dfrac{1}{\sqrt{2}}\hat{i} - \dfrac{1}{\sqrt{2}}\hat{j}\right)$

The extensions (or compressions) in the springs are the components of $\vec{s}$ along their respective directions:
For spring C, the displacement is directly along its axis, so it is compressed by $x$. The restoring force is $F_C = kx$ directed towards the origin.
For spring A, the extension is the magnitude of the component of $\vec{s}$ along $\hat{i}$, which is $x \cos 45^\circ = \dfrac{x}{\sqrt{2}}$. The restoring force is $F_A = \dfrac{kx}{\sqrt{2}}$ along $\hat{i}$.
For spring B, the extension is the magnitude of the component of $\vec{s}$ along $\hat{j}$, which is $x \cos 45^\circ = \dfrac{x}{\sqrt{2}}$. The restoring force is $F_B = \dfrac{kx}{\sqrt{2}}$ along $\hat{j}$.

The net restoring force along the line of displacement (towards the origin) is the sum of the components of these forces along that line:
$F_{net} = F_C + F_A \cos 45^\circ + F_B \cos 45^\circ$
$F_{net} = kx + \left(\dfrac{kx}{\sqrt{2}}\right)\left(\dfrac{1}{\sqrt{2}}\right) + \left(\dfrac{kx}{\sqrt{2}}\right)\left(\dfrac{1}{\sqrt{2}}\right)$
$F_{net} = kx + \dfrac{kx}{2} + \dfrac{kx}{2} = 2kx$

The effective spring constant is $k_{eff} = 2k$.

The time period of oscillation is:
$T = 2\pi \sqrt{\dfrac{m}{k_{eff}}} = 2\pi \sqrt{\dfrac{m}{2k}}$

Answer: $2\pi \sqrt{\dfrac{m}{2k}}$