We can have more arguments to prove that PQRS is a rhombus, but, the argument that we will use here is:
Let's look at the first statement, we have
[tex]PT>QT[/tex]
That's not correct, it would just prove that QR/2 > PS/2,
[tex]PR=QS[/tex]
This statement implies
[tex]\begin{gathered} PR^2=QS^2 \\ \\ PS^2+SR^2=PQ^2+QR^2 \end{gathered}[/tex]
We cannot conclude that
[tex]PS=SR=PQ=QR[/tex]
The next statement is
[tex]PT=QT[/tex]
A rhombus can have different diagonals, and in fact they have. Then let's go to the next one
[tex]ST=QT[/tex]
That also not exactly says it's a rhombus, it's a pallelogram property.
[tex]\angle SPT=\angle QPT[/tex]
By doing that we have that the diagonal bissects the angle
That implies that the angle b is also bissect.
The last statment is
[tex]\angle PTQ=\angle STR[/tex]
That's literally the vertex angle, it's true always, not only in that case, therefore the only possible answer is
[tex]\angle SPT=\angle QPT[/tex]
Pro
