Up to a unitary isomorphism, a Hilbert space is uniquely determined by its dimension, and closed subspaces are Hilbert spaces in their own right. So $P(H)$ is the disjoint union, over all cardinal numbers $\alpha \leq \dim(H)$, of closed subspaces of dimension $\alpha$. And there as many of the latter as unitary operators on a Hilbert space of dimension $\alpha$.

So in finite dimension, $\mathrm{card}(P(H)) = \sum_{\alpha=1}^{\dim(H)} \mathrm{card}(U(\alpha)) = \mathrm{card}(U(\dim(H))) = \mathrm{card}(\mathbb{R})$.

In infinite dimension, the unitary group $U(H)$ of $B(H)$ injects into the C*-algebra $B(H)$ of bounded operators on $H$. Also, any element of $B(H)$ is a linear combination of four unitaries, so there is a surjection $U(H)^4 \times \mathbb{C}^4 \to B(H)$. Hence $\mathrm{card}(P(H))=\mathrm{card}(U(H))=\mathrm{card}(B(H))=\max(\dim(H),\mathrm{card}(\mathbb{R}))$.

So regardless of dimension, $P(H)$ has cardinality $\max(\dim(H), 2^{\aleph_0})$.

Any Boolean subalgebra of $P(H)$ is contained in a maximal one, and those are all induced as the powerset of a choice of orthonormal basis. Hence the cardinality of the set of all Boolean subalgebras of $P(H)$ is $\mathrm{card}(U(H))$ (actually equal to $\mathrm{card}(P(H))$).

Up to a unitary isomorphism, a Hilbert space is uniquely determined by its dimension, and closed subspaces are Hilbert spaces in their own right. So $P(H)$ is the disjoint union, over all cardinal numbers $\alpha \leq \dim(H)$, of closed subspaces of dimension $\alpha$. And there as many of the latter as unitary operators on a Hilbert space of dimension $\alpha$.

So in finite dimension, $\mathrm{card}(P(H)) = \sum_{\alpha=1}^{\dim(H)} \mathrm{card}(U(\alpha)) = \mathrm{card}(U(\dim(H))) = \mathrm{card}(\mathbb{R})$.

In infinite dimension, the unitary group $U(H)$ of $B(H)$ injects into the C*-algebra $B(H)$ of bounded operators on $H$. Also, any element of $B(H)$ is a linear combination of four unitaries, so there is a surjection $U(H)^4 \times \mathbb{C}^4 \to B(H)$. Hence $\mathrm{card}(P(H))=\mathrm{card}(U(H))=\mathrm{card}(B(H))=\max(\dim(H),\mathrm{card}(\mathbb{R}))$.

So regardless of dimension, $P(H)$ has cardinality $\max(\dim(H), 2^{\aleph_0})$.

Any Boolean subalgebra of $P(H)$ is contained in a maximal one, and those are all induced as the powerset of a choice of orthonormal basis. Hence the cardinality of the set of all Boolean subalgebras of $P(H)$ is $2^{2^{\mathrm{card}(P(H))}}$.

Up to a unitary isomorphism, a Hilbert space is uniquely determined by its dimension, and closed subspaces are Hilbert spaces in their own right. So $P(H)$ is the disjoint union, over all cardinal numbers $\alpha \leq \dim(H)$, of closed subspaces of dimension $\alpha$. And there as many of the latter as unitary operators on a Hilbert space of dimension $\alpha$.

So in finite dimension, $\mathrm{card}(P(H)) = \sum_{\alpha=1}^{\dim(H)} \mathrm{card}(U(\alpha)) = \mathrm{card}(U(\dim(H))) = \mathrm{card}(\mathbb{R})$.

In infinite dimension, the unitary group $U(H)$ of $B(H)$ injects into the C*-algebra $B(H)$ of bounded operators on $H$. Also, any element of $B(H)$ is a linear combination of four unitaries, so there is a surjection $U(H)^4 \times \mathbb{C}^4 \to B(H)$. Hence $\mathrm{card}(U(H))=\mathrm{card}(B(H))=\dim(H)$, and so $\mathrm{card}(P(H)) = \dim(H)$.

Any Boolean subalgebra of $P(H)$ is contained in a maximal one, and those are all induced as the powerset of a choice of orthonormal basis. Hence the cardinality of the set of all Boolean subalgebras of $P(H)$ is $\mathrm{card}(U(H))$ (actually equal to $\mathrm{card}(P(H))$).

Up to a unitary isomorphism, a Hilbert space is uniquely determined by its dimension, and closed subspaces are Hilbert spaces in their own right. So $P(H)$ is the disjoint union, over all cardinal numbers $\alpha \leq \dim(H)$, of closed subspaces of dimension $\alpha$. And there as many of the latter as unitary operators on a Hilbert space of dimension $\alpha$.

So in finite dimension, $\mathrm{card}(P(H)) = \sum_{\alpha=1}^{\dim(H)} \mathrm{card}(U(\alpha)) = \mathrm{card}(U(\dim(H))) = \mathrm{card}(\mathbb{R})$.

In infinite dimension, the unitary group $U(H)$ of $B(H)$ injects into the C*-algebra $B(H)$ of bounded operators on $H$. Also, any element of $B(H)$ is a linear combination of four unitaries, so there is a surjection $U(H)^4 \times \mathbb{C}^4 \to B(H)$. Hence $\mathrm{card}(P(H))=\mathrm{card}(U(H))=\mathrm{card}(B(H))=\max(\dim(H),\mathrm{card}(\mathbb{R}))$.

So regardless of dimension, $P(H)$ has cardinality $\max(\dim(H), 2^{\aleph_0})$.

Any Boolean subalgebra of $P(H)$ is contained in a maximal one, and those are all induced as the powerset of a choice of orthonormal basis. Hence the cardinality of the set of all Boolean subalgebras of $P(H)$ is $\mathrm{card}(U(H))$ (actually equal to $\mathrm{card}(P(H))$).