On the diagram $M$ and $P$ are the midpoints of the edges $BC$ and $AB$ of the regular triangular prism $ABCA_1B_1C_1$.
Which of the following lines is perpendicular to the plane $(BCC_1B_1):$ $AB;\quad PM;\quad A_1M;\quad PC_1?$
I am not really sure about the most correct way to approach the problem, but I was able to eliminate two of the answers.
The line $AB$ is perpendicular to $BB_1$ as $ABB_1A_1$ is a rectangle, but it cannot be parallel to $BC$ as $\measuredangle ABC=60^\circ$. Therefore $AB$ cannot be perpendicular to the plane $(BCC_1B_1)$. If that was the case, it would be perpendicular to all lines in that plane.
I think that the angle between $PM$ and $(BCC_1B_1)$ is the same as the angle between $AC$ and $(BCC_1B_1)$. Is this actually true? But $AC$ is not perpendicular to $(BCC_1B_1)$ as $\measuredangle ACB=60^\circ$.
I don't see what to do with the other two lines. I noted that as $AM\perp BC$, then $A_1M\perp BC$, but that was it.
How to find if the other two lines are perpendicular to the given plane?
Your reasoning about the perpendicularity of the first two lines and the plane $(BCC_1B_1)$ is correct. For the third case suppose that $A_1M\bot (BCC_1B_1)$. Then $A_1M\bot MM_1$, where $M_1$ is middlepoint of $B_1C_1$. Consider triangle $AMM_1$. This triangle is a right triangle with right angle $\angle A_1M_1M=90^{\circ}$. So, then $\angle A_1MM_1<90^{\circ}$ and we get a contradiction. For the fourth case we will use the same argument and consider right triangle $PCC_1$ with right angle $\angle PCC_1=90^{\circ}$. Then $\angle PC_1C<90^{\circ}$ and therefore $PC_1$ is not perpendicular to $CC_1$ and hence, not perpendicular to the plane $(BCC_1B)$.