Let $X,H$ be Hilbert spaces and let $f\colon H \to L(H,X)$, where $L(H,X)$ is the space of linear bounded operators. In this case it is natural to define the adjoint function $f^*\colon H \to L(X,H)(=L(X^*,H^*))$ by $$f^* (x)=[f(x)]^* \quad \forall x \in H,$$ where $*$ denotes the adjoint of the operator $f(x) \in L(H,X).$
Clearly if $f$ is continuous in the operator topology, we have that $f^*$ is continuous in the operator topology, i.e. as $|x- x_0|_H\to 0$ $$\|f^*(x)-f^*(x_0)\|_{L(X,H)}=\|f(x)-f(x_0)\|_{L(H,X)} \to 0.$$
However, assume that $f$ is strongly continuous, i.e. $$\lim_{|x-x_0|_H\to 0}|f(x)\eta-f(x_0)\eta|_X=0 \quad \forall \eta \in H. $$ Is then $f^*$ strongly continuous in this case? I.e. we should have $$\lim_{|x-x_0|_H\to 0}|f^*(x)y-f^*(x_0)y|_X=0 \quad \forall y \in X$$
No. Let $H = X = l^2(\mathbb{N})$ and $v$ be the adjoint of the unilateral shift. We define a function $g: \mathbb{R} \to L(H)$ first by,
$$g(t) = \begin{cases} 0 &, \, \mathrm{if} \, t \leq 0\\ \frac{\frac{1}{n}-t}{\frac{1}{n}-\frac{1}{n+1}}v^{n+1}+\frac{t-\frac{1}{n+1}}{\frac{1}{n}-\frac{1}{n+1}}v^n &, \, \mathrm{if} \, \frac{1}{n+1} \leq t < \frac{1}{n}, \, n \in \mathbb{N}_+\\ v &, \, \mathrm{if} \, t \geq 1 \end{cases}$$
Since $v^n \to 0$ strongly as $n \to \infty$, it is easy to verify that $g$ is strongly continuous. Now simply let $f(h) = g(\mathrm{Re}\langle e_1, h \rangle)$. Clearly $f$ is still strongly continuous. But $f^\ast$ is not strongly continuous. Indeed, $\frac{1}{n}e_1 \to 0$, but $f^\ast(\frac{1}{n}e_1) = g(\frac{1}{n})^\ast = (v^n)^\ast$, which does not converge strongly (to $f^\ast(0) = 0$ or any other operator).