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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 $\mathbb{P}^3$.

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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 $\mathbb{P}^3$.

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More generally, the moduli space of algebraic K3 surfaces is a countable union on 19 dimensional subvarieties of the 20 dimensional moduli space of complex K3 surfaces, and exactly one of those components corresponds to K3 surfaces that embed as smooth quartic surfaces in $\mathbb{P}^3$.

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