Solution:

The answer is (A). Let $T_1$, $T_2$, $T_3$ be the points at which the tangents from $P$ touch the two circles, as in the figure. Observe that, by well-known properties of tangents, the angles $\widehat{O_1 T_1 P}$, $\widehat{O_1 T_2 P}$, $\widehat{O_2 T_2 P}$, $\widehat{O_2 T_3 P}$ are right angles. Note also that the segments $P T_1$, $P T_2$, $P T_3$ all have the same length: $\overline{P T_1}=\overline{P T_2}$ because they are the two tangent segments from $P$ to the same circle, and $\overline{P T_2}=\overline{P T_3}$ by the same reasoning, applied to the other circle.

The triangles $P T_1 O_1$, $P T_2 O_1$, $P T_2 O_2$, $P T_3 O_2$ are then all congruent, since each of them has one side equal to $1~\mathrm{cm}$ (the radii of the circles), one side equal to the length $\overline{P T_1}=\overline{P T_2}=\overline{P T_3}$, and the angle between them equal to $90^\circ$. Moreover, the angle $\widehat{T_2 P O_1}$, given the symmetry of the figure, is half of $\widehat{O_2 P O_1}=60^\circ$, hence $\widehat{T_2 P O_1}=\widehat{T_1 P O_1}=30^\circ$. In the triangle $P T_1 O_1$ we then have angles of $30^\circ$, $60^\circ$, $90^\circ$, which allows us to compute $\overline{P O_1}=2 \overline{T_1 O_1}=2~\mathrm{cm}$, $\overline{P T_1}=\frac{\sqrt{3}}{2} \overline{P O_1}=\sqrt{3}~\mathrm{cm}$ and hence

also the area of the triangle $P O_1 T_1$, equal to $\frac{1}{2} \overline{P T_1} \cdot \overline{T_1 O_1}=\frac{1}{2} \cdot 1~\mathrm{cm} \cdot \sqrt{3}~\mathrm{cm}=\frac{\sqrt{3}}{2}~\mathrm{cm}^2$.

The portion of the circle contained in the union of the triangles $P T_1 O_1$ and $P T_2 O_1$ is a circular segment with central angle $\widehat{T_1 O_1 T_2}=180^\circ-\widehat{T_2 P T_1}=120^\circ=\frac{1}{3} 360^\circ$, hence this area equals $\frac{1}{3} \pi(1~\mathrm{cm})^2$. The complement of this circular segment within the circle is in turn a circular segment, with central angle $240^\circ$ (and hence area $\frac{2}{3} \pi~\mathrm{cm}^2$).

The total area of the heart is then given by the area of the circular segments with endpoints $T_2 T_1$ and $T_3 T_2$ and angle $240^\circ$ (total area $2 \cdot \frac{2}{3} \pi~\mathrm{cm}^2$), plus the area of four triangles congruent to $P T_1 O_1$ (namely $P T_1 O_1$, $P T_2 O_1$, $P T_2 O_2$, $P T_3 O_2$), which as already seen each have area $\frac{\sqrt{3}}{2}~\mathrm{cm}^2$. The answer is therefore $\frac{4}{3} \pi+4 \frac{\sqrt{3}}{2}=\frac{4}{3} \pi+2 \sqrt{3}$.