= Solution
Here <functional calculus convergence> means that for every $f\in C(\mathbb T)$,
$$
\boxed{
\langle f(A_n)P_nv,P_nw\rangle
\longrightarrow
\langle f(A)v,w\rangle
\qquad(v,w\in\mathcal H).}
$$
Embed $\mathcal H_n$ in $\mathcal H$ and write $\widetilde A_n=P_n^*A_nP_n$. The hypothesis gives $\widetilde A_n\rightharpoonup A$ in the <weak operator topology>. Since $A_n$ and $A$ are <unitary operators>,
$$
\|\widetilde A_nv-Av\|^2
=\|P_nv\|^2+\|v\|^2
-2\operatorname{Re}\langle\widetilde A_nv,Av\rangle
\longrightarrow0.
$$
Thus $\widetilde A_n\to A$ in the <strong operator topology>. Applying the same argument to the adjoints gives $\widetilde A_n^*\to A^*$ strongly.
Products of uniformly bounded strongly convergent operators converge strongly, so for every integer $k$,
$$
P_n^*A_n^kP_n\longrightarrow A^k
$$
strongly, with negative $k$ interpreted through adjoints. Therefore convergence holds for every <Laurent polynomial>. The <Stone-Weierstrass theorem> says that Laurent polynomials are uniformly dense in $C(\mathbb T)$. Since the continuous functional calculus is contractive, uniform approximation finishes the proof for every $f\in C(\mathbb T)$.
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