A bounded linear operator from $X ^*\rightarrow Y^*$ is weak*-weak* continuous iff It is the adjoint of some Bounded linear operator $X\rightarrow Y$

The result is not true if $Y$ is not complete.

For instance take $X=Y\subset\ell^1$ be $$ X=Y=\{x\in\ell^1:\ \exists n_0:\ n\geq n_0\implies x(n)=0\}. $$ Because $X$ and $Y$ are dense in $\ell^1$, we have $X^*=Y^*=\ell^\infty$.

Define $S:Y^*\to X^*$, that is $S:\ell^\infty\to\ell^\infty$ by $$ Sw=\big(\sum_n\frac{w(n)}{n^2},0,0,\ldots\big). $$ If $w_j\to0$ weak$^*$, this means that $\sum_nw_j(n)y(n)\to0$ for all $y\in Y$. In particular $\sum_n\frac{w_j(n)}{n^2}\to0$, and it follows $S$ is weak$^*$-weak$^*$ continuous.

If we had $S=T^*$, with $T\in \mathcal L(X,Y)$ this would mean that, for each $w\in\ell^\infty$ and $x\in X$, $$ (Sw)x=w(Tx). $$ This translates to $$ \sum_n\frac{w(n)x(1)}{n^2}=\sum_nw(n)\,(Tx)(n). $$ As this should work for all $w\in\ell^\infty$, it follows that we need $$ Tx=\bigg(\frac{x(1)}{n^2}\bigg)_n. $$ But then $Tx\not\in Y$ for any nonzero $x$, and so $T\not\in \mathcal L(X,Y)$.