### Part (a)
1. Assume the contrary: Suppose a ring $A$ with property $(P)$ has an odd number of elements. Let $|A| = n$ and $|R| = m$, where $R$ is a non-trivial subgroup of $(A, +)$ and $(U(A), \cdot) \cong (R, +)$. Since $R$ is a subgroup of $(A, +)$, $m$ divides $n$.

2. **Odd order of $R$**: Since $|A|$ is odd, $m$ must also be odd. Consequently, $|U(A)| = m$ is odd.

3. Properties of units: In a ring, the set of units $U(A)$ is a group under multiplication. If $|U(A)|$ is odd, then $-1$ cannot be a unit unless $-1 = 1$, which implies $2 = 0$ in $A$.

4. Contradiction: If $2 = 0$ in $A$, then $A$ has characteristic 2, which implies $|A|$ is even (since the characteristic of a finite ring divides its order). This contradicts our assumption that $|A|$ is odd.

5. **Structure of $(A, +)$**: Write $(A, +) = C_{k_1} \oplus \cdots \oplus C_{k_s}$ where $C_{k_i}$ are cyclic groups of order $k_i \geq 2$. Since $|A|$ is odd, all $k_i$ must be odd.

6. Even order: If all $k_i$ are odd, then $|A|$ is a product of odd numbers, which is odd. However, we previously derived that $|A|$ must be even, leading to a contradiction.

Thus, the number of elements in $A$ must be even.

### Part (b)
1. Constructing rings: Consider the ring $A = \mathbb{Z}/(2 \cdot 3^n) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/3^n$ for $n \geq 1$.

2. **Units of $A$**: The units of $A$ are given by $U(A) \cong (\mathbb{Z}/3^n)^*$. The group $(\mathbb{Z}/3^n)^*$ is cyclic of order $2 \cdot 3^{n-1}$.

3. **Embedding into $(A, +)$**: Since $2 \cdot 3^{n-1}$ divides $2 \cdot 3^n$, we can embed $(\mathbb{Z}/3^n)^*$ into $(A, +)$.

4. Distinct orders: For each $n$, the ring $A$ has order $2 \cdot 3^n$. Since $n$ can be any positive integer, there are infinitely many distinct orders of rings with property $(P)$.