Algebraic $K3$ surface means the $K3$ surface admits an ample line bundle. So the question is equivalent to asking whether every algebraic $K3$ surface can be embedded in $\mathbb{P}^3$.



Is every algebraic $K3$ surface a quartic surface?
No. Consider a K3 surface with a polarization of degree 2 and with Picard rank 1. Since the tautological line bundle on $\mathbb{P}^3$ pulls back to a degree 4 line bundle, it follows that such a K3 surface cannot be embedded in $\mathbb{P}^3$.


