The type of hybridization and the magnetic property of $$[MnCl_6]^{3-}$$ are :

Step 1 : Find the oxidation state of Mn in $$[MnCl_6]^{3-}$$.
Each chloride ligand is $$Cl^-$$, carrying charge $$-1$$.
Let the oxidation state of Mn be $$x$$.
Therefore $$x + 6(-1) = -3 \;\Rightarrow\; x - 6 = -3 \;\Rightarrow\; x = +3$$.
So the central metal is $$Mn^{3+}$$.

Step 2 : Write the ground-state electronic configuration of $$Mn^{3+}$$.
Atomic number of Mn is 25: $$Mn : [Ar]\,3d^5\,4s^2$$.
For $$Mn^{3+}$$ we remove two $$4s$$ electrons and one $$3d$$ electron: $$Mn^{3+} : [Ar]\,3d^4$$.

Step 3 : Decide whether the complex is inner-orbital or outer-orbital.
Chloride $$(Cl^-)$$ is a weak-field ligand in the spectro-chemical series.
Weak-field ligands do not cause pairing of $$3d$$ electrons.
Hence the complex remains high-spin, and the metal uses the outer (fourth) shell orbitals for hybridisation.

Step 4 : Determine the hybridisation.
An octahedral complex requires six hybrid orbitals.
Because no pairing occurs in $$3d$$, all five $$3d$$ orbitals are occupied by four unpaired electrons and one empty orbital, leaving no two vacant inner $$d$$ orbitals.
Therefore Mn must use $$4s$$, three $$4p$$, and two $$4d$$ orbitals: $$sp^3d^2$$ (outer-orbital octahedral).

Step 5 : Calculate the number of unpaired electrons.
Configuration of $$Mn^{3+}$$ remains $$3d^4$$ with weak-field ligands: $$\uparrow\,\uparrow\,\uparrow\,\uparrow$$ (four unpaired electrons).
Hence the complex is paramagnetic with four unpaired electrons.

Conclusion : The complex $$[MnCl_6]^{3-}$$ exhibits $$sp^3d^2$$ hybridisation and is paramagnetic with four unpaired electrons.

Therefore the correct choice is Option B.