A radioactive element ${ }_{92}^{242} \mathrm{X}$ emits two $\alpha$ particles, one electron and two positrons. The product nucleus is represented by ${ }_{\mathrm{P}}^{234} \mathrm{Y}$. The value of P is

Determining the atomic number $P$ follows from applying the rules of radioactive decay to the transformation ${ }_{92}^{242} \mathrm{X} \to { }_{P}^{234} \mathrm{Y}$.

An $\alpha$ decay reduces the mass number by 4 and the atomic number by 2. After two $\alpha$ emissions, the mass number becomes $242 - 2 \times 4 = 234$ and the atomic number becomes $92 - 2 \times 2 = 88$.

A $\beta^-$ decay leaves the mass number unchanged and increases the atomic number by 1. After this emission, the atomic number becomes $88 + 1 = 89$.

A $\beta^+$ decay leaves the mass number unchanged and decreases the atomic number by 1. After two such decays, the atomic number becomes $89 - 2 \times 1 = 87$.

Since the final nucleus is ${ }_{P}^{234} \mathrm{Y}$, where the mass number 234 matches our result, the atomic number is $P = 87$.

$\boxed{87}$