Boron in $BCl_3$ has only three bonding pairs and no lone pair around it, since boron contributes just three valence electrons to form three B-Cl bonds. With three electron domains and nothing else, VSEPR predicts $sp^2$ hybridisation and a flat, trigonal planar shape with 120 degree bond angles.
Because boron has an empty p orbital in this arrangement, $BCl_3$ is electron deficient and readily accepts a lone pair, making it a good Lewis acid. Ammonia, on the other hand, carries a lone pair on nitrogen and is a Lewis base, so the two combine through a coordinate bond, nitrogen donating its lone pair into boron's empty orbital.
Once that coordinate bond forms, boron is surrounded by four bonding pairs, the three original B-Cl bonds plus the new B to N bond, with no lone pair of its own. Four electron domains around a central atom means $sp^3$ hybridisation, which gives a tetrahedral shape.
So $BCl_3$ is trigonal planar and $BCl_3\cdot NH_3$ is tetrahedral, matching option (1).