The molar solubility(s) of zirconium phosphate with molecular formula $$(Zr^{4+})_{3}(PO_{4}^{3-})_{4}$$ is given by relation :

Zirconium phosphate has the formula $$(Zr^{4+})_3(PO_4^{3-})_4$$.

The dissociation equilibrium is:

$$Zr_3(PO_4)_4 \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$$

If the molar solubility is $$s$$, then:

$$[Zr^{4+}] = 3s$$ and $$[PO_4^{3-}] = 4s$$

The solubility product expression is:

$$K_{sp} = [Zr^{4+}]^3 \cdot [PO_4^{3-}]^4 = (3s)^3 \cdot (4s)^4$$

$$K_{sp} = 27s^3 \cdot 256s^4 = 6912 \cdot s^7$$

Solving for $$s$$:

$$s = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}}$$

The correct answer is Option B: $$\left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}}$$.