What is the isotope formed when 232/90 Th undergoes alpha decay?

alpha decay is when an element releases a 4/2 He particle

1. B

I'm not sure about the other one

Th-232  ==>  He-4 + Ra-228  (both A and B)

214............0......214
.......Bi  ==>  e + ......Pb
 83.............1........82

An alpha particle contains 2 protons and 2 neutrons, so daughter product produced by alpha decay has an atomic number (Z) two less than the parent, and an atomic mass that is 4 units less than the parent.  

For 232/90Th, we have a daughter that has (Z/mass) = (232-4)/(90-2) = 228/88Ra.

Emission of a positron converts a proton to a neutron.  (A neutrino is also emitted, but this has no effect on the identity of the daughter nucleus.)  This process does not change the atomic mass, but it decreases the atomic number (Z) by 1.  If 214/83Bi emitted a positron, it would become a 214/82Pb nucleus.  

In reality 214/83Bi does not decay by positron emission, it mostly decays by "regular" beta decay (emission of an electron and an antineutrino), so 214/82Bi typically decays to 214/84Po.  A small fraction of the time (about 0.02% of the decays) it decays by alpha emission, resulting in a 210/81Tl nucleus.

Tell your teacher to stop writing questions that don't correspond to reality.