I am not 100% clear what you are asking, but I will answer according to two interpretations:
a) Suppose that $F$, a metrizable TVS, is the strong dual of $E$, a locally convex TVS. Need $F$ be complete?
The answer to this question is no, by the following counterexample, where we obtain an incomplete normed space as the strong dual of a locally convex space.
Take $E$ to be $B(\mathcal{H})$, the bounded operators on an an infinite dimensional Hilbert space $\mathcal{H}$. The topology on $B(\mathcal{H})$ we will use is the strong operator topology (SOT, i.e. pointwise convergence of Hilbert space norm). It is not hard to show that, as a vector space, $F = E'$ is $\mathrm{FinRank}(\mathcal{H})$, the finite-rank operators (see, for example, Takesaki's Theory of Operator Algebras I, Theorem II.2.6). To work out what the strong topology on $F$ is, we need to know what the bounded sets are for the SOT on $B(\mathcal{H})$.
To this end, we can use the completeness of $\mathcal{H}$ and apply the uniform boundedness principle (see Schaefer's Topological Vector Spaces, III.4.2), showing SOT-bounded sets are exactly equicontinuous sets, a.k.a. norm-bounded sets in $B(\mathcal{H})$. Therefore the strong dual topology on $F$ is that defined by the dual norm, which is the trace-class norm on $\mathrm{FinRank}(\mathcal{H})$. As $\mathcal{H}$ is infinite-dimensional, $F$ is not complete in this norm, as there are trace-class operators that are not of finite rank.
b) Suppose that $F$, a metrizable TVS, is the strong dual of $E$, a metrizable TVS. Need $F$ be complete?
The answer to this question is yes, and also that $F$ is a Banach space (or "Banachable" space, without choosing any norm in particular). This is a consequence of this answer, plus the fact that the dual of a normed space is a Banach space.