Preliminaries. Let $H$ be a Hilbert space. Consider bilinear operator
$$
\bigcirc: H\times H^{cc}\to\mathcal{F}(H):(x,y)\mapsto (z\mapsto \langle z,y\rangle x)
$$
where $H^{cc}$ is a complex conjgate Hilbert space and $\mathcal{F}(H)$ is a normed space of all finite rank operators with operator norm. One can check that $\bigcirc$ is bounded bilinear operator of norm one and its image is all rank one operators. For a given $x,y\in H$ this is straightforward to check that
$$
A(x\bigcirc y)B=A(x)\bigcirc B^*(y)\tag{1}
$$
Lemma Let $H$ be a Hilbert space and $(T_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $T\in\mathcal{B}(H)$. Then, $(T_n:n\in\mathbb{N})$ is norm bounded in $\mathcal{B}(H)$ by some constant $c_T$.
Proof. Since $(T_n:n\in\mathbb{N})$ strongly converges to $T$, then for all $x\in X$ the sequence $(T_n(x):n\in\mathbb{N})$ is norm bounded in $H$. By Banach-Steinhaus theorem the sequence $(T_n:n\in\mathbb{N})$ is norm bounded in $\mathcal{B}(H)$ by some constant $c_T>0$
Lemma. Let $H$ be a Hilbert space $A\in\mathcal{K}(H)$ and $(T_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $T\in\mathcal{B}(H)$, $(S_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $S\in\mathcal{B}(H)$ then so $(T_nAS_n^*:n\in\mathbb{N})$ converges to $TAS^*$ in the norm topology of $\mathcal{B}(H)$.
Proof. Let $x,y\in H$. Since $(T_n:n\in\mathbb{N})$ and $(S_n:n\in\mathbb{N})$ converges to $T$ and $S$ respectively in the strong operator topology, then $(T_n(x):n\in\mathbb{N})$ and $(S_n(y):n\in\mathbb{N})$ converge to $T(x)$ and $S(y)$ respectively in the norm topology of $H$. By properties of bilinear operator $\bigcirc$ we conclude that $(T_n(x)\bigcirc S_n(y):n\in\mathbb{N})$ converges to $T(x)\bigcirc S(y)$ in the norm topology of $\mathcal{B}(H)$. Therefore from $(1)$ we get that
$$
\lim\limits_{n\to\infty} T_n(x\bigcirc y) S_n^*=T(x\bigcirc y)S^*\tag{2}
$$
in the norm topology of $\mathcal{B}(H)$. Now take arbitrary $F\in\mathcal{F}(H)$. Since every finite rank operator is a sum of rank one operators, then there exists $(x_1,\ldots,x_k)\subset H$ and $(y_1,\ldots,y_k)\subset H$ such that $F=\sum_{i=1}^k x_k\bigcirc y_k$. So using $(2)$ we get
$$
\lim\limits_{n\to\infty} T_nF S_n^*
=\sum_{i=1}^k\lim\limits_{n\to\infty} T_n(x_k\bigcirc y_k) S_n^*
=\sum_{i=1}^k T(x_k\bigcirc y_k)S^*
=TFS^*\tag{3}
$$
in the norm topology of $\mathcal{B}(H)$.
Finally, consider compact operator $A\in\mathcal{K}(H)$. Since $H$ is a Hilbert space, then $A$ is limit of finite rank operators in the norm topology of $\mathcal{B}(H)$. Fix $\varepsilon>0$. From the previous note we have that there exist $F\in\mathcal{F}(H)$ and $D\in\mathcal{K}(H)$ such that $A=F+D$ and $\Vert D\Vert<\varepsilon$. Then using previous lemma we get
$$
\Vert T_n DS_n^*-TDS^*\Vert
\leq\Vert T_n DS_n^*\Vert+\Vert TDS^*\Vert
\leq\Vert T_n\Vert \Vert D\Vert \Vert S_n^*\Vert+\Vert T\Vert\Vert D\Vert \Vert S^*\Vert
\leq c_Tc_S \varepsilon +\Vert T\Vert\Vert S\Vert \varepsilon
$$
$$
0\leq\Vert T_n AS_n^*-T AS^*\Vert
\leq \Vert T_n FS_n^*-T FS^*\Vert+\Vert T_n DS_n^*-T DS^*\Vert\\
\leq \Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon
$$
Now we take $\limsup$ when $n\to\infty$ and use $(3)$ to get
$$
0\leq\limsup\limits_{n\to\infty}\Vert T_n AS_n^*-T AS^*\Vert
\leq \limsup\limits_{n\to\infty}\Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon\\
\leq \lim\limits_{n\to\infty}\Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon
\leq (c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon
$$
Since $\varepsilon>0$ is arbitrary from previous inequalities we conclude that $\lim\limits_{n\to\infty}\Vert T_n AS_n^*-T AS^*\Vert$ exists and equals to $0$. Hence we proved that
$$
\lim\limits_{n\to\infty} T_n AS_n^*=T AS^*
$$
in the norm topology of $\mathcal{B}(H)$.
Proposition Let $H$ be a Hilbert space and $P\in\mathcal{K}(H)$. Assume $\gamma:[0,1]\to\mathcal{B}(H):t\mapsto v_tpv_t^*$ is a path continuous in strong operator topology, then it is continuous in norm topology.
Proof Since $[0,1]$ is a first-countable topological space it is enough to show sequential continuity. For a given sequence $(t_n:n\in\mathbb{N})\subset[0,1]$ convrgent to $t\in[0,1]$ define $S_n=T_n=v_{t_n}$ and apply previous lemma to get that $v_{t_n}Pv_{t_n}^*$ converges to $v_tPv_t^*$ in norm. Hence $\gamma$ is norm continuous path.
