The bond order and magnetic behaviour of $$O_2^-$$ ion are, respectively:

The electronic configuration of one $$\text{O}$$ atom is $$1s^2\,2s^2\,2p^4$$ (i.e. six valence electrons).
Therefore the neutral $$\text{O}_2$$ molecule contains $$2\times6 = 12$$ valence electrons.

For molecules from $$\text{N}_2$$ to $$\text{Ne}_2$$ the molecular-orbital energy order is:
$$$\sigma(1s),\;\sigma^*(1s),\;\sigma(2s),\;\sigma^*(2s),\;\sigma(2p_z),\;\pi(2p_x)=\pi(2p_y),\;\pi^*(2p_x)=\pi^*(2p_y),\;\sigma^*(2p_z)$$$.

Filling these orbitals with 12 electrons (Hund’s rule + Pauli principle) gives neutral $$\text{O}_2$$ the configuration
$$$\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \sigma(2p_z)^2$$$ $$$\pi(2p_x)^2\, \pi(2p_y)^2\, \pi^*(2p_x)^1\, \pi^*(2p_y)^1$$$.

Bond order is defined as
$$$\text{Bond order} = \frac{(\text{number of bonding electrons}) - (\text{number of antibonding electrons})}{2}$$$.

For neutral $$\text{O}_2$$:
Bonding electrons $$= 8$$ (in $$\sigma(2p_z),\;\pi(2p_x),\;\pi(2p_y)$$).
Antibonding electrons $$= 4$$ (in $$\pi^*(2p_x),\;\pi^*(2p_y)$$).
Thus $$\text{B.O.} = \frac{8-4}{2} = 2$$.

Now consider $$\text{O}_2^-$$ (superoxide ion). It possesses one extra electron, so total valence electrons $$= 12 + 1 = 13$$.

This extra electron enters the next available orbital $$\pi^*(2p)$$, giving the configuration
$$\dots \pi^*(2p_x)^2\,\pi^*(2p_y)^1$$.

Counting again:
Bonding electrons remain $$8$$.
Antibonding electrons become $$5$$.

Hence
$$\text{Bond order} = \frac{8 - 5}{2} = \frac{3}{2} = 1.5$$.

Magnetic behaviour depends on unpaired electrons. In $$\text{O}_2^-$$ the orbitals contain one unpaired electron (in the singly-occupied $$\pi^*$$ orbital). Any species with at least one unpaired electron is paramagnetic.

Therefore the superoxide ion $$\text{O}_2^-$$ has
Bond order = 1.5, Paramagnetic behaviour.

Thus, correct choice: Option B.